Textbook - Areas of Parallelograms and Triangles Questions in English

(N/A) The figures $(i), (iii)$ and $(v)$ lie on the same base and between the same parallels.

Figure	Common base and two parallels
Fig. $(i)$	Common base: $DC$,Parallels: $DC$ and $AB$
Fig. $(iii)$	Common base: $QR$,Parallels: $QR$ and $PS$
Fig. $(v)$	Common base: $AD$,Parallels: $AD$ and $BQ$

(N/A) $(i)$ Since a rectangle is also a parallelogram,and both $ABCD$ and $EFCD$ are on the same base $DC$ and between the same parallels $DC$ and $EF$,therefore,$\text{ar}(ABCD) = \text{ar}(EFCD)$.
$(ii)$ From the result above,$\text{ar}(ABCD) = \text{ar}(EFCD)$.
Since $EFCD$ is a rectangle,its area is $\text{length} \times \text{breadth} = DC \times FC$.
Thus,$\text{ar}(ABCD) = DC \times FC$ $(1)$.
Since $AL \perp DC$ and $EF \parallel DC$,$AL$ is the height of the parallelogram. In the rectangle $AFCL$ (or by considering the parallel lines),we have $AL = FC$ $(2)$.
Substituting $(2)$ in $(1)$,we get $\text{ar}(ABCD) = DC \times AL$.

(N/A) Let $\Delta ABP$ and parallelogram $ABCD$ be on the same base $AB$ and between the same parallels $AB$ and $PC$.
You wish to prove that $\text{ar}(PAB) = \frac{1}{2} \text{ar}(ABCD)$.
Draw $BQ \parallel AP$ to obtain another parallelogram $ABQP$. Now,parallelograms $ABQP$ and $ABCD$ are on the same base $AB$ and between the same parallels $AB$ and $PC$.
Therefore,$\text{ar}(ABQP) = \text{ar}(ABCD)$ $(1)$.
But $\Delta PAB \cong \Delta BQP$ (Diagonal $PB$ divides parallelogram $ABQP$ into two congruent triangles).
So,$\text{ar}(PAB) = \text{ar}(BQP)$ $(2)$.
Therefore,$\text{ar}(PAB) = \frac{1}{2} \text{ar}(ABQP)$ [From $(2)$] $(3)$.
This gives $\text{ar}(PAB) = \frac{1}{2} \text{ar}(ABCD)$ [From $(1)$ and $(3)$].

(D) We are given that $ABCD$ is a parallelogram with $AE \perp DC$ and $CF \perp AD$.
Given $AB = 16 \, cm, AE = 8 \, cm$ and $CF = 10 \, cm$.
Since $ABCD$ is a parallelogram,opposite sides are equal,so $CD = AB = 16 \, cm$.
The area of a parallelogram is given by the product of its base and the corresponding height.
Area of parallelogram $ABCD = CD \times AE = 16 \, cm \times 8 \, cm = 128 \, cm^2$.
Also,the area of the same parallelogram can be calculated using base $AD$ and height $CF$.
Area of parallelogram $ABCD = AD \times CF$.
Substituting the known values,we get $128 = AD \times 10$.
Therefore,$AD = \frac{128}{10} = 12.8 \, cm$.

(N/A) Let us join $E$ and $G$.
If a triangle and a parallelogram are on the same base and between the same parallels,then the area of the triangle is equal to half the area of the parallelogram.
Since $E$ and $G$ are the mid-points of $AB$ and $CD$ respectively,$EG$ is parallel to $BC$ and $AD$.
Also,$\text{ar}(\text{parallelogram } EBCG) = \text{ar}(\text{parallelogram } AEGD) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD) \dots (1)$
Now,$\Delta EFG$ and parallelogram $EBCG$ are on the same base $EG$,and between the same parallels $EG$ and $BC$.
Therefore,$\text{ar}(\Delta EFG) = \frac{1}{2} \text{ar}(\text{parallelogram } EBCG) \dots (2)$
Similarly,$\text{ar}(\Delta EHG) = \frac{1}{2} \text{ar}(\text{parallelogram } AEGD) \dots (3)$
Adding $(2)$ and $(3)$,we get:
$\text{ar}(\Delta EFG) + \text{ar}(\Delta EHG) = \frac{1}{2} [\text{ar}(\text{parallelogram } EBCG) + \text{ar}(\text{parallelogram } AEGD)]$
$\Rightarrow \text{ar}(EFGH) = \frac{1}{2} [\text{ar}(\text{parallelogram } ABCD)]$
Thus,$\text{ar}(EFGH) = \frac{1}{2} \text{ar}(ABCD)$.

(N/A) Given: $ABCD$ is a parallelogram.
Since $ABCD$ is a parallelogram,we have $AB \parallel CD$ and $BC \parallel AD$.
Now,$\Delta APB$ and parallelogram $ABCD$ are on the same base $AB$ and between the same parallels $AB$ and $CD$.
Therefore,$\text{ar}(\Delta APB) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD) \quad \dots(1)$
Also,$\Delta BQC$ and parallelogram $ABCD$ are on the same base $BC$ and between the same parallels $BC$ and $AD$.
Therefore,$\text{ar}(\Delta BQC) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD) \quad \dots(2)$
From $(1)$ and $(2)$,we have:
$\text{ar}(\Delta APB) = \text{ar}(\Delta BQC)$

(N/A) $(i)$ We have a parallelogram $ABCD$,i.e.,$AB \parallel CD$ and $BC \parallel AD$.
Let us draw a line $EF \parallel AB$ passing through $P$,where $E$ lies on $AD$ and $F$ lies on $BC$.
Since $AB \parallel EF$,$\Delta APB$ and parallelogram $AEFB$ are on the same base $AB$ and between the same parallels $AB$ and $EF$.
Therefore,$\operatorname{ar}(\Delta APB) = \frac{1}{2} \operatorname{ar}(AEFB)$ ...... $(1)$
Also,$\Delta PCD$ and parallelogram $CDEF$ are on the same base $CD$ and between the same parallels $CD \parallel EF$.
Therefore,$\operatorname{ar}(\Delta PCD) = \frac{1}{2} \operatorname{ar}(CDEF)$ ...... $(2)$
Adding $(1)$ and $(2)$,we have:
$\operatorname{ar}(\Delta APB) + \operatorname{ar}(\Delta PCD) = \frac{1}{2} [\operatorname{ar}(AEFB) + \operatorname{ar}(CDEF)]$
$\operatorname{ar}(\Delta APB) + \operatorname{ar}(\Delta PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(ii)$ Let us draw a line $GH \parallel AD$ passing through $P$,where $G$ lies on $CD$ and $H$ lies on $AB$.
$\Delta APD$ and parallelogram $ADGH$ are on the same base $AD$ and between the same parallels $AD$ and $GH$.
Therefore,$\operatorname{ar}(\Delta APD) = \frac{1}{2} \operatorname{ar}(ADGH)$ ...... $(3)$
Similarly,$\Delta PBC$ and parallelogram $BCGH$ are on the same base $BC$ and between the same parallels $BC$ and $GH$.
Therefore,$\operatorname{ar}(\Delta PBC) = \frac{1}{2} \operatorname{ar}(BCGH)$ ...... $(4)$
Adding $(3)$ and $(4)$,we have:
$\operatorname{ar}(\Delta APD) + \operatorname{ar}(\Delta PBC) = \frac{1}{2} [\operatorname{ar}(ADGH) + \operatorname{ar}(BCGH)]$
$\operatorname{ar}(\Delta APD) + \operatorname{ar}(\Delta PBC) = \frac{1}{2} \operatorname{ar}(ABCD)$ ...... $(5)$
From $(i)$,we know $\operatorname{ar}(\Delta APB) + \operatorname{ar}(\Delta PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$ ...... $(6)$
Comparing $(5)$ and $(6)$,we get:
$\operatorname{ar}(\Delta APD) + \operatorname{ar}(\Delta PBC) = \operatorname{ar}(\Delta APB) + \operatorname{ar}(\Delta PCD)$

(N/A) We have two parallelograms $PQRS$ and $ABRS$ and a point $X$ on $BR$.
$(i)$ To prove that $ar(PQRS) = ar(ABRS)$:
Since parallelogram $PQRS$ and parallelogram $ABRS$ are on the same base $RS$ and between the same parallels $RS$ and $PB$,their areas are equal.
Therefore,$ar(PQRS) = ar(ABRS)$.
$(ii)$ To prove that $ar(AXS) = 1/2 \, ar(PQRS)$:
Since $\Delta AXS$ and parallelogram $ABRS$ are on the same base $AS$ and between the same parallels $AS$ and $BR$,the area of the triangle is half the area of the parallelogram.
Therefore,$ar(AXS) = 1/2 \, ar(ABRS)$.
Since $ar(PQRS) = ar(ABRS)$ (from part $i$),
Substituting this into the equation,we get:
$ar(AXS) = 1/2 \, ar(PQRS)$.

(N/A) The farmer has a field in the form of a parallelogram $PQRS$,and a point $A$ is situated on $RS$.
Let us join $AP$ and $AQ$.
Obviously,the field is divided into three parts,i.e.,$\Delta APS$,$\Delta PAQ$,and $\Delta QAR$. These parts are triangular in shape.
Since $\Delta PAQ$ and parallelogram $PQRS$ are on the same base $PQ$ and between the same parallels $PQ$ and $RS$:
$\therefore \text{ar}(\Delta PAQ) = \frac{1}{2} \text{ar}(\text{parallelogram } PQRS) \dots(1)$
$\Rightarrow \text{ar}(\text{parallelogram } PQRS) - \text{ar}(\Delta PAQ) = \text{ar}(\text{parallelogram } PQRS) - \frac{1}{2} \text{ar}(\text{parallelogram } PQRS)$
$\Rightarrow [\text{ar}(\Delta APS) + \text{ar}(\text{QAR})] = \frac{1}{2} \text{ar}(\text{parallelogram } PQRS) \dots(2)$
From $(1)$ and $(2)$,we have:
$\text{ar}(\Delta PAQ) = \text{ar}(\Delta APS) + \text{ar}(\Delta QAR)$
Thus,the farmer can sow wheat in $\Delta PAQ$ and pulses in the combined area of $\Delta APS$ and $\Delta QAR$,or vice versa.

(N/A) Let $\Delta ABC$ be a triangle and let $AD$ be one of its medians,where $D$ is the midpoint of $BC$.
We wish to show that $\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.
Since the formula for the area of a triangle involves an altitude,let us draw $AN \perp BC$.
Now,$\operatorname{ar}(ABD) = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times BD \times AN$.
Since $AD$ is a median,$BD = CD$.
Substituting $BD = CD$ in the expression for $\operatorname{ar}(ABD)$,we get:
$\operatorname{ar}(ABD) = \frac{1}{2} \times CD \times AN$.
Since $\frac{1}{2} \times CD \times AN$ is the formula for the area of $\Delta ACD$,we have:
$\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.
Thus,a median of a triangle divides it into two triangles of equal areas.

(N/A) Observe the figure carefully.
$\Delta BAC$ and $\Delta EAC$ lie on the same base $AC$ and between the same parallel lines $AC$ and $BE$.
Therefore,$ar(\Delta BAC) = ar(\Delta EAC)$ (Triangles on the same base and between the same parallels are equal in area).
Now,add $ar(\Delta ADC)$ to both sides of the equation:
$ar(\Delta BAC) + ar(\Delta ADC) = ar(\Delta EAC) + ar(\Delta ADC)$
From the figure,$ar(\Delta BAC) + ar(\Delta ADC) = ar(ABCD)$ and $ar(\Delta EAC) + ar(\Delta ADC) = ar(\Delta ADE)$.
Therefore,$ar(ABCD) = ar(\Delta ADE)$.

(N/A) We have a $\Delta ABC$ such that $AD$ is a median.
Since a median of a triangle divides it into two triangles of equal areas,
$\text{ar} (\Delta ABD) = \text{ar} (\Delta ADC) \quad \dots(1)$
Similarly,in $\Delta EBC$,$ED$ is a median.
Therefore,$\text{ar} (\Delta EBD) = \text{ar} (\Delta ECD) \quad \dots(2)$
Subtracting equation $(2)$ from equation $(1)$,we get:
$\text{ar} (\Delta ABD) - \text{ar} (\Delta EBD) = \text{ar} (\Delta ADC) - \text{ar} (\Delta ECD)$
$\Rightarrow \text{ar} (\Delta ABE) = \text{ar} (\Delta ACE)$.

(N/A) We have a $\Delta ABC$ and its median $AD$.
Since a median divides the triangle into two triangles equal in area,
$\therefore \operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(\Delta ABC) \quad \dots(1)$
Now,in $\Delta ABD$,$BE$ is a median because $E$ is the mid-point of $AD$.
$\therefore \operatorname{ar}(\Delta BED) = \frac{1}{2} \operatorname{ar}(\Delta ABD) \quad \dots(2)$
From $(1)$ and $(2)$,we have:
$\operatorname{ar}(\Delta BED) = \frac{1}{2} \left[ \frac{1}{2} \operatorname{ar}(\Delta ABC) \right]$
$\Rightarrow \operatorname{ar}(\Delta BED) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$

(N/A) Let $ABCD$ be a parallelogram such that its diagonals $AC$ and $BD$ intersect at $O$.
Since the diagonals of a parallelogram bisect each other,we have $AO = OC$ and $BO = OD$.
Draw $CE \perp BD$.
Now,$\text{ar}(\Delta BOC) = \frac{1}{2} \times BO \times CE$ and $\text{ar}(\Delta DOC) = \frac{1}{2} \times OD \times CE$.
Since $BO = OD$,it follows that $\text{ar}(\Delta BOC) = \text{ar}(\Delta DOC) \quad ... (1)$.
Similarly,by drawing perpendiculars from $A$ to $BD$,we can show that $\text{ar}(\Delta AOD) = \text{ar}(\Delta AOB) \quad ... (2)$.
Also,considering $\Delta ABD$ and $\Delta CBD$ on the same base $BD$ and between the same parallels $AD \parallel BC$,we know that $\text{ar}(\Delta ABD) = \text{ar}(\Delta CBD)$.
Since $\text{ar}(\Delta ABD) = \text{ar}(\Delta AOB) + \text{ar}(\Delta AOD)$ and $\text{ar}(\Delta CBD) = \text{ar}(\Delta BOC) + \text{ar}(\Delta DOC)$,we have $\text{ar}(\Delta AOB) + \text{ar}(\Delta AOD) = \text{ar}(\Delta BOC) + \text{ar}(\Delta DOC)$.
Using the relations derived,all four triangles have equal area: $\text{ar}(\Delta AOB) = \text{ar}(\Delta BOC) = \text{ar}(\Delta COD) = \text{ar}(\Delta DOA)$.
Thus,the diagonals of a parallelogram divide it into four triangles of equal area.

(N/A) We have $\Delta ABC$ and $\Delta ABD$ on the same base $AB$.
Since $CD$ is bisected at $O$,we have $CO = DO$.
In $\Delta ACD$,$AO$ is a median because it divides the side $CD$ into two equal parts $(CO = DO)$.
Therefore,$\operatorname{ar}(\Delta AOC) = \operatorname{ar}(\Delta AOD)$ (Since a median of a triangle divides it into two triangles of equal areas) $(1)$.
Similarly,in $\Delta BCD$,$BO$ is a median.
Therefore,$\operatorname{ar}(\Delta BOC) = \operatorname{ar}(\Delta BOD)$ $(2)$.
Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\Delta AOC) + \operatorname{ar}(\Delta BOC) = \operatorname{ar}(\Delta AOD) + \operatorname{ar}(\Delta BOD)$
$\Rightarrow \operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta ABD)$.

(N/A) We have: $\triangle ABC$ such that the mid-points of $BC, CA$ and $AB$ are respectively $D, E$ and $F$.
To prove that $BDEF$ is a parallelogram.
In $\triangle ABC$,$E$ and $F$ are the mid-points of $AC$ and $AB$ respectively.
By the Mid-point Theorem,the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
Therefore,$EF || BC$ and $EF = \frac{1}{2} BC$.
Since $D$ is the mid-point of $BC$,we have $BD = \frac{1}{2} BC$.
Thus,$EF || BD$ and $EF = BD$.
Since $BDEF$ is a quadrilateral in which one pair of opposite sides ($EF$ and $BD$) is parallel and equal in length,$BDEF$ is a parallelogram.

(N/A) Given: $D, E, F$ are mid-points of sides $BC, CA, AB$ respectively in $\Delta ABC$.
To prove: $\operatorname{ar}(DEF) = \frac{1}{4} \operatorname{ar}(ABC)$.
Proof:
$1$. Since $D$ and $F$ are mid-points of $BC$ and $AB$ respectively,by the Mid-point Theorem,$DF \parallel AC$ and $DF = \frac{1}{2} AC = AE$. Thus,$AFDE$ is a parallelogram.
$2$. Similarly,$BDEF$ and $CDFE$ are parallelograms.
$3$. In parallelogram $BDEF$,the diagonal $DF$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta BDF) = \operatorname{ar}(\Delta DEF)$.
$4$. In parallelogram $AFDE$,the diagonal $DE$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta AFE) = \operatorname{ar}(\Delta DEF)$.
$5$. In parallelogram $CDFE$,the diagonal $DE$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta CDE) = \operatorname{ar}(\Delta DEF)$.
$6$. Therefore,$\operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta AFE) + \operatorname{ar}(\Delta BDF) + \operatorname{ar}(\Delta CDE) + \operatorname{ar}(\Delta DEF)$.
$7$. Substituting the equal areas: $\operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) = 4 \operatorname{ar}(\Delta DEF)$.
$8$. Hence,$\operatorname{ar}(\Delta DEF) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.

(N/A) Given that $D, E$ and $F$ are the mid-points of sides $BC, CA$ and $AB$ respectively.
By the Mid-point Theorem,$FE \parallel BC$ and $FE = \frac{1}{2} BC = BD$.
Since $FE \parallel BD$ and $FE = BD$,$BDEF$ is a parallelogram.
Similarly,$AFDE$ and $FDCE$ are also parallelograms.
In parallelogram $BDEF$,the diagonal $DE$ divides it into two triangles of equal area,so $\operatorname{ar}(\Delta BDF) = \operatorname{ar}(\Delta DEF)$.
Similarly,in parallelogram $AFDE$,$\operatorname{ar}(\Delta AFE) = \operatorname{ar}(\Delta DEF)$,and in parallelogram $FDCE$,$\operatorname{ar}(\Delta DEF) = \operatorname{ar}(\Delta EDC)$.
Thus,$\operatorname{ar}(\Delta AFE) = \operatorname{ar}(\Delta BDF) = \operatorname{ar}(\Delta DEF) = \operatorname{ar}(\Delta EDC) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
Now,$\operatorname{ar}(BDEF) = \operatorname{ar}(\Delta BDF) + \operatorname{ar}(\Delta DEF) = \frac{1}{4} \operatorname{ar}(\Delta ABC) + \frac{1}{4} \operatorname{ar}(\Delta ABC) = \frac{2}{4} \operatorname{ar}(\Delta ABC) = \frac{1}{2} \operatorname{ar}(\Delta ABC)$.

(A) We have a quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ intersect at $O$.
We are given that $OB = OD$ and $AB = CD$.
Let us draw $DE \perp AC$ and $BF \perp AC$.
$(i)$ To prove that $ar(\Delta DOC) = ar(\Delta AOB)$:
In $\Delta DEO$ and $\Delta BFO$,we have:
$DO = BO$ (Given)
$\angle DOE = \angle BOF$ (Vertically opposite angles)
$\angle DEO = \angle BFO = 90^\circ$ (By construction)
Therefore,$\Delta DEO \cong \Delta BFO$ ($AAS$ congruence rule).
This implies $DE = BF$ and $ar(\Delta DEO) = ar(\Delta BFO)$ $(1)$.
Now,in $\Delta DEC$ and $\Delta BFA$,we have:
$\angle DEC = \angle BFA = 90^\circ$
$DE = BF$ (From above)
$DC = BA$ (Given)
Therefore,$\Delta DEC \cong \Delta BFA$ ($RHS$ congruence rule).
This implies $ar(\Delta DEC) = ar(\Delta BFA)$ $(2)$.
Adding $(1)$ and $(2)$,we get:
$ar(\Delta DEO) + ar(\Delta DEC) = ar(\Delta BFO) + ar(\Delta BFA)$
$ar(\Delta DOC) = ar(\Delta AOB)$.
$(ii)$ To prove that $ar(DCB) = ar(ACB)$:
Since $ar(\Delta DOC) = ar(\Delta AOB)$,
Adding $ar(\Delta BOC)$ on both sides,we have:
$ar(\Delta DOC) + ar(\Delta BOC) = ar(\Delta AOB) + ar(\Delta BOC)$
$ar(\Delta DCB) = ar(\Delta ACB)$.
$(iii)$ To prove that $DA \parallel CB$:
Since $\Delta DCB$ and $\Delta ACB$ are on the same base $CB$ and have equal areas,they must lie between the same parallels.
Therefore,$DA \parallel CB$. Since $DA \parallel CB$ and $AB = CD$ (with $OB=OD$),$ABCD$ is a parallelogram.

(N/A) Given: In $\Delta ABC$,$D$ and $E$ are points on $AB$ and $AC$ such that $\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$.
To prove: $DE \parallel BC$.
Proof:
We are given that $\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$.
Both triangles $\Delta DBC$ and $\Delta EBC$ lie on the same base $BC$.
We know that if two triangles have the same base and equal areas,they lie between the same parallels.
Therefore,the line segment $DE$ must be parallel to the base $BC$.
Hence,$DE \parallel BC$.

(N/A) We have a $\Delta ABC$ such that $XY || BC$,$BE || AC$,and $CF || AB$.
Since $XY || BC$ and $BE || AC$ (given $BE || AC$ and $E$ lies on $XY$),we have $BCYE$ as a parallelogram.
Now,the parallelogram $BCYE$ and $\Delta ABE$ are on the same base $BE$ and between the same parallels $BE$ and $AC$.
Therefore,$\operatorname{ar}(\Delta ABE) = \frac{1}{2} \operatorname{ar}(BCYE) \quad -(1)$
Again,since $CF || AB$ and $XY || BC$,we have $BCFX$ as a parallelogram.
Now,$\Delta ACF$ and parallelogram $BCFX$ are on the same base $CF$ and between the same parallels $AB$ and $CF$.
Therefore,$\operatorname{ar}(\Delta ACF) = \frac{1}{2} \operatorname{ar}(BCFX) \quad -(2)$
Also,parallelogram $BCFX$ and parallelogram $BCYE$ are on the same base $BC$ and between the same parallels $BC$ and $EF$.
Therefore,$\operatorname{ar}(BCFX) = \operatorname{ar}(BCYE) \quad -(3)$
From equations $(1)$,$(2)$,and $(3)$,we get:
$\operatorname{ar}(\Delta ABE) = \operatorname{ar}(\Delta ACF)$.

(A) We have a parallelogram $ABCD$. $AB$ is produced to $P$.
$CB$ is produced to $Q$ and parallelogram $PBQR$ is completed.
Let us join $AC$ and $PQ$.
Since $ABCD$ is a parallelogram and $AC$ is its diagonal,
$\text{ar}(\Delta ABC) = \frac{1}{2} \text{ar}(ABCD) \quad \dots(1)$
Since $PBQR$ is a parallelogram and $PQ$ is its diagonal,
$\text{ar}(\Delta PBQ) = \frac{1}{2} \text{ar}(PBQR) \quad \dots(2)$
Since $\Delta ACQ$ and $\Delta APQ$ are on the same base $AQ$ and between the same parallels $AQ$ and $CP$,
$\text{ar}(\Delta ACQ) = \text{ar}(\Delta APQ)$
Subtracting $\text{ar}(\Delta ABQ)$ from both sides:
$\text{ar}(\Delta ACQ) - \text{ar}(\Delta ABQ) = \text{ar}(\Delta APQ) - \text{ar}(\Delta ABQ)$
$\text{ar}(\Delta ABC) = \text{ar}(\Delta PBQ) \quad \dots(3)$
From $(1), (2),$ and $(3)$,we get:
$\frac{1}{2} \text{ar}(ABCD) = \frac{1}{2} \text{ar}(PBQR)$
Therefore,$\text{ar}(ABCD) = \text{ar}(PBQR)$.

(N/A) We have a trapezium $ABCD$ with $AB || DC$. The diagonals $AC$ and $BD$ intersect at $O$.
Since triangles on the same base and between the same parallels have equal areas:
$\because \Delta ABD$ and $\Delta ABC$ are on the same base $AB$ and between the same parallels $AB$ and $DC$,
$\therefore \operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta ABC)$.
Subtracting $\operatorname{ar}(\Delta AOB)$ from both sides,we get:
$\operatorname{ar}(\Delta ABD) - \operatorname{ar}(\Delta AOB) = \operatorname{ar}(\Delta ABC) - \operatorname{ar}(\Delta AOB)$
$\Rightarrow \operatorname{ar}(\Delta AOD) = \operatorname{ar}(\Delta BOC)$.

(N/A) We have a pentagon $ABCDE$ in which $BF \parallel AC$ and $DC$ is produced to $F$.
$(i)$ To prove $ar(\Delta ACB) = ar(\Delta ACF)$:
Since triangles on the same base and between the same parallels are equal in area.
$\because \Delta ACB$ and $\Delta ACF$ are on the same base $AC$ and between the same parallels $AC$ and $BF$.
$\therefore ar(\Delta ACB) = ar(\Delta ACF)$.
$(ii)$ Since $ar(\Delta ACB) = ar(\Delta ACF)$:
Adding $ar(AEDC)$ to both sides,we get:
$ar(\Delta ACB) + ar(AEDC) = ar(\Delta ACF) + ar(AEDC)$
$\Rightarrow ar(ABCDE) = ar(AEDF)$.

(N/A) Let the quadrilateral plot be $ABCD$. Let $E$ be the point where the diagonal $AC$ intersects the boundary of the portion to be taken. To implement the proposal,we draw a line through $D$ parallel to $AC$,which meets the extended side $BC$ at point $F$.
Now,consider $\Delta DAF$ and $\Delta DCF$. These triangles are on the same base $DF$ and between the same parallel lines $AC$ and $DF$.
Therefore,$\text{ar}(\Delta DAF) = \text{ar}(\Delta DCF)$.
Subtracting $\text{ar}(\Delta DEF)$ from both sides,we get:
$\text{ar}(\Delta DAF) - \text{ar}(\Delta DEF) = \text{ar}(\Delta DCF) - \text{ar}(\Delta DEF)$
$\Rightarrow \text{ar}(\Delta ADE) = \text{ar}(\Delta CEF)$.
This means the area of the triangular portion $\Delta ADE$ (which the Panchayat takes) is equal to the area of the triangular portion $\Delta CEF$ (which is given to Itwaari). By adding $\Delta CEF$ to the remaining part of the plot,Itwaari gets a new triangular plot $\Delta ABF$.
To verify the area: $\text{ar}(\Delta ABF) = \text{ar}(ABCE) + \text{ar}(\Delta CEF)$.
Since $\text{ar}(\Delta CEF) = \text{ar}(\Delta ADE)$,we have:
$\text{ar}(\Delta ABF) = \text{ar}(ABCE) + \text{ar}(\Delta ADE) = \text{ar}(\text{quadrilateral } ABCD)$.
Thus,the total area remains the same.

(N/A) We have a trapezium $ABCD$ such that $AB || DC$.
$XY || AC$ meets $AB$ at $X$ and $BC$ at $Y$.
Let us join $CX$.
$\because AB || DC$ [Given]
$\therefore \Delta ADX$ and $\Delta ACX$ are on the same base $AX$ and between the same parallels $AB$ and $DC$.
$\therefore \operatorname{ar}(\Delta ADX) = \operatorname{ar}(\Delta ACX) \quad \dots(1)$
$\because AC || XY$
$\therefore \Delta ACX$ and $\Delta ACY$ are on the same base $AC$ and between the same parallels $AC$ and $XY$.
$\therefore \operatorname{ar}(\Delta ACX) = \operatorname{ar}(\Delta ACY) \quad \dots(2)$
From $(1)$ and $(2)$,we have
$\operatorname{ar}(\Delta ADX) = \operatorname{ar}(\Delta ACY)$

(N/A) Given: $AP || BQ || CR$.
Step $1$: Since $BQ || CR$,$\Delta BCQ$ and $\Delta BQR$ are on the same base $BQ$ and between the same parallels $BQ$ and $CR$.
Therefore,$\operatorname{ar}(\Delta BCQ) = \operatorname{ar}(\Delta BQR)$ --- $(1)$
Step $2$: Since $AP || BQ$,$\Delta ABQ$ and $\Delta PBQ$ are on the same base $BQ$ and between the same parallels $AP$ and $BQ$.
Therefore,$\operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta PBQ)$ --- $(2)$
Step $3$: Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\Delta BCQ) + \operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta BQR) + \operatorname{ar}(\Delta PBQ)$
From the figure,$\operatorname{ar}(\Delta BCQ) + \operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta AQC)$ and $\operatorname{ar}(\Delta BQR) + \operatorname{ar}(\Delta PBQ) = \operatorname{ar}(\Delta PBR)$.
Thus,$\operatorname{ar}(\Delta AQC) = \operatorname{ar}(\Delta PBR)$.
Hence proved.

(N/A) We have a quadrilateral $ABCD$ and its diagonals $AC$ and $BD$ intersect at $O$ such that $\operatorname{ar}(\Delta AOD) = \operatorname{ar}(\Delta BOC)$.
Now,$\operatorname{ar}(\Delta AOD) = \operatorname{ar}(\Delta BOC)$ [Given].
Adding $\operatorname{ar}(\Delta AOB)$ to both sides,we have:
$\operatorname{ar}(\Delta AOD) + \operatorname{ar}(\Delta AOB) = \operatorname{ar}(\Delta BOC) + \operatorname{ar}(\Delta AOB)$
$\Rightarrow \operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta ABC)$.
Since these two triangles $\Delta ABD$ and $\Delta ABC$ are on the same base $AB$ and have equal areas,they must lie between the same parallel lines.
Therefore,$AB \parallel DC$.
Since $ABCD$ is a quadrilateral having a pair of opposite sides parallel,$ABCD$ is a trapezium.

(N/A) We have $ar(\Delta DRC) = ar(\Delta DPC)$ [Given].
Since they are on the same base $DC$ and have equal areas,they must lie between the same parallels.
$\Rightarrow DC \parallel RP$.
Since one pair of opposite sides of quadrilateral $DCPR$ is parallel,$DCPR$ is a trapezium.
Again,we have $ar(\Delta BDP) = ar(\Delta ARC)$ $(1)$.
Also,$ar(\Delta DPC) = ar(\Delta DRC)$ $(2)$.
Subtracting $(2)$ from $(1)$,we get:
$[ar(\Delta BDP) - ar(\Delta DPC)] = [ar(\Delta ARC) - ar(\Delta DRC)]$
$\Rightarrow ar(\Delta BDC) = ar(\Delta ADC)$.
Since they are on the same base $DC$ and have equal areas,they must lie between the same parallels.
$\Rightarrow AB \parallel DC$.
Since one pair of opposite sides of quadrilateral $ABCD$ is parallel,$ABCD$ is a trapezium.

(N/A) Given: Parallelogram $ABCD$ and rectangle $ABEF$ are on the same base $AB$ and $ar(ABCD) = ar(ABEF)$.
$1$. Since they are on the same base $AB$ and have equal areas,their heights must be equal. Let the height be $h = AF = BE$.
$2$. In a rectangle,the opposite sides are equal,so $AB = EF$. In a parallelogram,opposite sides are equal,so $AB = CD$. Thus,$CD = EF$.
$3$. The perimeter of rectangle $ABEF = 2(AB + AF)$.
$4$. The perimeter of parallelogram $ABCD = 2(AB + BC)$.
$5$. In the right-angled triangle $\triangle BEC$,$BC$ is the hypotenuse and $BE$ is one of the sides. Therefore,$BC > BE$.
$6$. Adding $AB$ to both sides of the inequality $BC > BE$,we get $AB + BC > AB + BE$.
$7$. Multiplying by $2$,we get $2(AB + BC) > 2(AB + BE)$.
$8$. Since $AB = EF$,the perimeter of the rectangle is $2(AB + BE) = AB + EF + BE + AF$.
$9$. Thus,the perimeter of the parallelogram $ABCD$ is greater than the perimeter of the rectangle $ABEF$.

(N/A) Let us draw $AF$ perpendicular to $BC$ such that $AF$ is the height of $\Delta ABD$,$\Delta ADE$,and $\Delta AEC$.
We know that the area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Therefore,$ar(ABD) = \frac{1}{2} \times BD \times AF$.
Similarly,$ar(ADE) = \frac{1}{2} \times DE \times AF$.
And $ar(AEC) = \frac{1}{2} \times EC \times AF$.
Since it is given that $BD = DE = EC$,we can substitute these values.
Thus,$\frac{1}{2} \times BD \times AF = \frac{1}{2} \times DE \times AF = \frac{1}{2} \times EC \times AF$.
This implies $ar(ABD) = ar(ADE) = ar(AEC)$.
Yes,since the altitudes (heights) of all the triangles are the same and their bases are equal,the areas are equal. Therefore,Budhia can use this result to divide her land into three equal parts.

(N/A) Given: $ABCD$,$DCFE$,and $ABFE$ are parallelograms.
$1$. Since $ABCD$ is a parallelogram,its opposite sides are parallel and equal. Therefore,$AD = BC$ and $AD \parallel BC$. Also,$AB \parallel DC$....$(1)$
$2$. Since $DCFE$ is a parallelogram,its opposite sides are parallel and equal. Therefore,$DC \parallel EF$....$(2)$
$3$. From $(1)$ and $(2)$,we have $AB \parallel DC$ and $DC \parallel EF$,which implies $AB \parallel EF$.
$4$. Now,consider $\Delta ADE$ and $\Delta BCF$.
- We have $AD = BC$ (opposite sides of parallelogram $ABCD$).
- We have $DE = CF$ (opposite sides of parallelogram $DCFE$).
- We have $AE = BF$ (opposite sides of parallelogram $ABFE$).
- Thus,$\Delta ADE \cong \Delta BCF$ by $SSS$ congruence criterion.
$5$. Since the triangles are congruent,their areas must be equal.
- Therefore,$\operatorname{ar}(ADE) = \operatorname{ar}(BCF)$.

(N/A) We have a parallelogram $ABCD$ and $AD = CQ$.
Let us join $AC$. We know that triangles on the same base and between the same parallels are equal in area.
Since $\Delta ADC$ and $\Delta ABC$ are on the same base $AC$ and between the same parallels $AB$ and $DC$,we have $\operatorname{ar}(\Delta ADC) = \operatorname{ar}(\Delta ABC)$.
Also,since $AD \parallel BQ$ and $AD = CQ$,we consider $\Delta ADC$ and $\Delta QDC$. However,a more direct approach is to use the property that triangles on the same base and between the same parallels are equal in area.
Consider $\Delta ADC$ and $\Delta ACQ$. Since $AD \parallel BQ$ (as $AD \parallel BC$),$\Delta ADC$ and $\Delta ACQ$ are between the same parallels $AD$ and $BQ$. But they do not share the same base.
Let us use the property: $\operatorname{ar}(\Delta ADC) = \operatorname{ar}(\Delta ABC)$.
Adding $\operatorname{ar}(\Delta APC)$ to both sides is not helpful. Instead,consider $\Delta ADQ$ and $\Delta ACQ$. Since $AD \parallel QC$,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ADC)$.
Actually,the standard proof is:
$1$. $\operatorname{ar}(\Delta ADC) = \operatorname{ar}(\Delta ABC)$ (Triangles on same base $AC$ and between parallels $AB \parallel DC$)
$2$. Since $AD \parallel BQ$,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$ (Triangles on same base $AQ$ and between parallels $AD \parallel BQ$ is not correct here).
Correct approach:
Since $AD \parallel BQ$,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$ is false.
Correct logic: $\operatorname{ar}(\Delta ADC) = \operatorname{ar}(\Delta ACQ)$ is false.
Let's use: $\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$ is not true.
Actually,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$ is true if $AD \parallel QC$. Yes,$AD \parallel BC$,so $AD \parallel QC$. Thus,$\operatorname{ar}(\Delta ADQ) = \operatorname{ar}(\Delta ACQ)$.
Subtracting $\operatorname{ar}(\Delta APD)$ from both sides:
$\operatorname{ar}(\Delta ADQ) - \operatorname{ar}(\Delta APD) = \operatorname{ar}(\Delta ACQ) - \operatorname{ar}(\Delta APD)$
$\operatorname{ar}(\Delta DPQ) = \operatorname{ar}(\Delta APC)$
Since $\operatorname{ar}(\Delta APC) = \operatorname{ar}(\Delta BPC)$ (triangles on same base $PC$ and between parallels $AB \parallel DC$),
Therefore,$\operatorname{ar}(\Delta DPQ) = \operatorname{ar}(\Delta BPC)$.

(A) Let the side of equilateral triangle $ABC$ be $2x$. Then $BC = 2x$. Since $D$ is the mid-point of $BC$,$BD = DC = x$. Since $BDE$ is an equilateral triangle,$BD = DE = BE = x$.
$(i)$ $\operatorname{ar}(ABC) = \frac{\sqrt{3}}{4} (2x)^2 = \sqrt{3}x^2$. $\operatorname{ar}(BDE) = \frac{\sqrt{3}}{4} x^2$. Thus,$\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$.
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \times \text{base} \times \text{height}$. Since $BE$ and $BA$ are sides of equilateral triangles,$\triangle BAE$ has base $BE=x$ and height equal to the altitude of $\triangle ABC$,which is $\sqrt{3}x$. $\operatorname{ar}(BAE) = \frac{1}{2} \times x \times \sqrt{3}x = \frac{\sqrt{3}}{2}x^2 = 2 \operatorname{ar}(BDE)$. Hence,$\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$.
$(iii)$ Since $D$ is the mid-point of $BC$,$\triangle BEC$ and $\triangle ABC$ share the same altitude from $A$ to $BC$. Thus,$\operatorname{ar}(BEC) = \frac{1}{2} \operatorname{ar}(ABC)$,which implies $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$.
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(ABE) - \operatorname{ar}(ABF)$. Using properties of triangles on the same base and between same parallels,it can be shown that $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$.
$(v)$ Since $F$ is the centroid of $\triangle ABC$ (or by geometric ratios),$\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$.
$(vi)$ By calculating the areas relative to the total area,$\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$.

(N/A) We have a quadrilateral $ABCD$ such that its diagonals $AC$ and $BD$ intersect at $P$. Let us draw $AM \perp BD$ and $CN \perp BD$.
$ar(\Delta APB) = \frac{1}{2} \times BP \times AM$
$ar(\Delta CPD) = \frac{1}{2} \times DP \times CN$
$ar(\Delta APB) \times ar(\Delta CPD) = (\frac{1}{2} \times BP \times AM) \times (\frac{1}{2} \times DP \times CN)$
$= \frac{1}{4} \times BP \times DP \times AM \times CN$ ... $(1)$
Similarly,
$ar(\Delta APD) = \frac{1}{2} \times DP \times AM$
$ar(\Delta BPC) = \frac{1}{2} \times BP \times CN$
$ar(\Delta APD) \times ar(\Delta BPC) = (\frac{1}{2} \times DP \times AM) \times (\frac{1}{2} \times BP \times CN)$
$= \frac{1}{4} \times BP \times DP \times AM \times CN$ ... $(2)$
From $(1)$ and $(2)$,we get
$ar(\Delta APB) \times ar(\Delta CPD) = ar(\Delta APD) \times ar(\Delta BPC)$

(N/A) Given: In $\Delta ABC$,$P$ is the mid-point of $AB$,$Q$ is the mid-point of $BC$,and $R$ is the mid-point of $AP$.
To prove: $\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (ARC)$.
Proof:
$1$. In $\Delta ABQ$,$P$ is the mid-point of $AB$. Since the median divides a triangle into two triangles of equal area,$PQ$ is a median of $\Delta ABQ$.
Therefore,$\operatorname{ar} (APQ) = \frac{1}{2} \operatorname{ar} (ABQ)$.
$2$. In $\Delta APQ$,$R$ is the mid-point of $AP$. Thus,$QR$ is a median of $\Delta APQ$.
Therefore,$\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (APQ) = \frac{1}{2} \times (\frac{1}{2} \operatorname{ar} (ABQ)) = \frac{1}{4} \operatorname{ar} (ABQ)$.
$3$. Since $AQ$ is a median of $\Delta ABC$,$\operatorname{ar} (ABQ) = \frac{1}{2} \operatorname{ar} (ABC)$.
Substituting this,$\operatorname{ar} (PRQ) = \frac{1}{4} \times (\frac{1}{2} \operatorname{ar} (ABC)) = \frac{1}{8} \operatorname{ar} (ABC)$.
$4$. Now,consider $\Delta ARC$. Since $R$ is the mid-point of $AP$,$CR$ is a median of $\Delta APC$.
Therefore,$\operatorname{ar} (ARC) = \frac{1}{2} \operatorname{ar} (APC)$.
$5$. Since $CP$ is a median of $\Delta ABC$,$\operatorname{ar} (APC) = \frac{1}{2} \operatorname{ar} (ABC)$.
Therefore,$\operatorname{ar} (ARC) = \frac{1}{2} \times (\frac{1}{2} \operatorname{ar} (ABC)) = \frac{1}{4} \operatorname{ar} (ABC)$.
$6$. Comparing the results: $\operatorname{ar} (PRQ) = \frac{1}{8} \operatorname{ar} (ABC)$ and $\operatorname{ar} (ARC) = \frac{1}{4} \operatorname{ar} (ABC)$.
Clearly,$\frac{1}{8} \operatorname{ar} (ABC) = \frac{1}{2} \times (\frac{1}{4} \operatorname{ar} (ABC))$.
Thus,$\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (ARC)$.

(N/A) Given: $P$ is the mid-point of $AB$,$Q$ is the mid-point of $BC$,and $R$ is the mid-point of $AP$.
$1$. Since $P$ is the mid-point of $AB$,$CP$ is a median of $\Delta ABC$. Therefore,$\operatorname{ar}(\Delta PBC) = \frac{1}{2} \operatorname{ar}(\Delta ABC)$.
$2$. In $\Delta PBC$,$PQ$ is a median (since $Q$ is the mid-point of $BC$). Therefore,$\operatorname{ar}(\Delta PBQ) = \operatorname{ar}(\Delta PQC) = \frac{1}{2} \operatorname{ar}(\Delta PBC) = \frac{1}{2} \times \frac{1}{2} \operatorname{ar}(\Delta ABC) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
$3$. In $\Delta APQ$,$RQ$ is a median (since $R$ is the mid-point of $AP$). Therefore,$\operatorname{ar}(\Delta RQC)$ is not directly related,but consider $\Delta RQC$ as part of $\Delta AQC$.
Alternatively,$\operatorname{ar}(\Delta RQC) = \operatorname{ar}(\Delta RQP) + \operatorname{ar}(\Delta PQC)$.
Since $R$ is the mid-point of $AP$,$\operatorname{ar}(\Delta RQP) = \frac{1}{2} \operatorname{ar}(\Delta APQ)$.
Since $P$ is the mid-point of $AB$,$\operatorname{ar}(\Delta APQ) = \frac{1}{2} \operatorname{ar}(\Delta ABQ) = \frac{1}{2} \times \frac{1}{2} \operatorname{ar}(\Delta ABC) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
So,$\operatorname{ar}(\Delta RQP) = \frac{1}{2} \times \frac{1}{4} \operatorname{ar}(\Delta ABC) = \frac{1}{8} \operatorname{ar}(\Delta ABC)$.
$4$. Thus,$\operatorname{ar}(\Delta RQC) = \operatorname{ar}(\Delta RQP) + \operatorname{ar}(\Delta PQC) = \frac{1}{8} \operatorname{ar}(\Delta ABC) + \frac{1}{4} \operatorname{ar}(\Delta ABC) = \frac{1+2}{8} \operatorname{ar}(\Delta ABC) = \frac{3}{8} \operatorname{ar}(\Delta ABC)$.

(N/A) Given: $P$ is the mid-point of $AB$,$Q$ is the mid-point of $BC$,and $R$ is the mid-point of $AP$.
$1$. In $\Delta ABC$,$AQ$ is a median (since $Q$ is the mid-point of $BC$). Therefore,$\text{ar}(\Delta ABQ) = \frac{1}{2} \text{ar}(\Delta ABC)$.
$2$. In $\Delta ABQ$,$PQ$ is a median (since $P$ is the mid-point of $AB$). Therefore,$\text{ar}(\Delta PBQ) = \frac{1}{2} \text{ar}(\Delta ABQ) = \frac{1}{2} \times (\frac{1}{2} \text{ar}(\Delta ABC)) = \frac{1}{4} \text{ar}(\Delta ABC)$.
$3$. In $\Delta ABC$,$AC$ is a side. Consider $\Delta ARC$. Since $R$ is the mid-point of $AP$ and $P$ is the mid-point of $AB$,$AR = RP = PB = \frac{1}{4} AB$.
$4$. Alternatively,using the property of medians: In $\Delta APC$,$CR$ is a median. Thus,$\text{ar}(\Delta ARC) = \frac{1}{2} \text{ar}(\Delta APC)$.
$5$. Since $P$ is the mid-point of $AB$,$\text{ar}(\Delta APC) = \frac{1}{2} \text{ar}(\Delta ABC)$.
$6$. Therefore,$\text{ar}(\Delta ARC) = \frac{1}{2} \times (\frac{1}{2} \text{ar}(\Delta ABC)) = \frac{1}{4} \text{ar}(\Delta ABC)$.
Thus,$\text{ar}(\Delta PBQ) = \text{ar}(\Delta ARC) = \frac{1}{4} \text{ar}(\Delta ABC)$.

(N/A) We have a right-angled $\Delta ABC$ such that $BCED$,$ACFG$,and $ABMN$ are squares on its sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ is drawn such that it meets $BC$ at $Y$.
To prove that $\Delta MBC \cong \Delta ABD$:
In $\Delta ABD$ and $\Delta MBC$,we have:
$AB = MB$ (Sides of the square $ABMN$)
$BD = BC$ (Sides of the square $BCED$)
$\angle MBA = 90^\circ$ and $\angle CBD = 90^\circ$ (Angles of squares)
Therefore,$\angle MBA = \angle CBD = 90^\circ$.
Adding $\angle ABC$ to both sides:
$\angle MBA + \angle ABC = \angle CBD + \angle ABC$
$\angle MBC = \angle ABD$
Thus,by the $SAS$ (Side-Angle-Side) congruence criterion:
$\Delta MBC \cong \Delta ABD$.

(N/A) To prove: $\operatorname{ar}(BYXD) = 2 \operatorname{ar}(\Delta MBC)$
$1$. Consider $\Delta ABD$ and $\Delta MBC$. In these triangles,$AB = MB$ (sides of square $ABMN$) and $BD = BC$ (sides of square $BCED$). Also,$\angle ABD = \angle ABC + 90^\circ$ and $\angle MBC = \angle ABC + 90^\circ$. Thus,$\angle ABD = \angle MBC$. By $SAS$ congruence,$\Delta ABD \cong \Delta MBC$. Therefore,$\operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta MBC)$.
$2$. Parallelogram $BYXD$ and $\Delta ABD$ are on the same base $BD$ and between the same parallels $BD$ and $AX$. Therefore,$\operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(BYXD)$.
$3$. Substituting the result from step $1$ into step $2$: $\operatorname{ar}(\Delta MBC) = \frac{1}{2} \operatorname{ar}(BYXD)$.
$4$. Hence,$\operatorname{ar}(BYXD) = 2 \operatorname{ar}(\Delta MBC)$.

(N/A) To prove: $\operatorname{ar}(BYXD) = \operatorname{ar}(ABMN)$.
$1$. Consider $\Delta MBC$ and $\Delta ABD$. We have $MB = AB$ (sides of square $ABMN$) and $BC = BD$ (sides of square $BCED$). Also,$\angle MBC = \angle MBA + \angle ABC = 90^\circ + \angle ABC$ and $\angle ABD = \angle ABC + \angle CBD = \angle ABC + 90^\circ$. Thus,$\angle MBC = \angle ABD$.
$2$. By $SAS$ congruence,$\Delta MBC \cong \Delta ABD$. Therefore,$\operatorname{ar}(\Delta MBC) = \operatorname{ar}(\Delta ABD)$.
$3$. Since $\Delta ABD$ and rectangle $BYXD$ are on the same base $BD$ and between the same parallels $AX$ and $BD$,$\operatorname{ar}(BYXD) = 2 \operatorname{ar}(\Delta ABD)$.
$4$. Since square $ABMN$ and $\Delta MBC$ are on the same base $MB$ and between the same parallels $MB$ and $NC$,$\operatorname{ar}(ABMN) = 2 \operatorname{ar}(\Delta MBC)$.
$5$. Since $\operatorname{ar}(\Delta ABD) = \operatorname{ar}(\Delta MBC)$,it follows that $2 \operatorname{ar}(\Delta ABD) = 2 \operatorname{ar}(\Delta MBC)$.
$6$. Therefore,$\operatorname{ar}(BYXD) = \operatorname{ar}(ABMN)$.

(N/A) To prove: $\Delta FCB \cong \Delta ACE$
In $\Delta FCB$ and $\Delta ACE$:
$1$. $FC = AC$ (Sides of the same square $ACFG$)
$2$. $CB = CE$ (Sides of the same square $BCED$)
$3$. $\angle FCA = \angle BCE = 90^\circ$ (Angles of squares)
Adding $\angle ACB$ to both sides:
$\angle FCA + \angle ACB = \angle BCE + \angle ACB$
$\angle FCB = \angle ACE$
Therefore,by the $SAS$ (Side-Angle-Side) congruence criterion:
$\Delta FCB \cong \Delta ACE$

(N/A) To prove that $\operatorname{ar}(CYXE) = 2 \operatorname{ar}(FCB)$:
$1$. Consider $\triangle ACE$ and square $BCED$. Both lie on the same base $CE$ and between the same parallel lines $AX$ and $CE$ (since $AX \perp DE$ and $BCED$ is a square,$BC \parallel DE$,thus $AX \parallel CE$ is not correct,but rather the altitude of the triangle from $A$ to $DE$ is $AX$,and $CYXE$ is a rectangle formed by the projection). Actually,$\triangle ACE$ and rectangle $CYXE$ are on the same base $CE$ and between the same parallels $AX$ and $CE$ is incorrect. Let's re-evaluate: $\triangle ACE$ and rectangle $CYXE$ are on the same base $CE$ and between the same parallels $AX$ and $CE$ is not the standard approach.
$2$. Correct approach: $\triangle ACE$ and rectangle $CYXE$ are on the same base $CE$ and between the same parallels $AX$ and $CE$. Thus,$\operatorname{ar}(CYXE) = 2 \operatorname{ar}(\triangle ACE)$.
$3$. Now,consider $\triangle ACE$ and $\triangle FCB$. In $\triangle ACE$ and $\triangle FCB$:
- $AC = FC$ (sides of square $ACFG$)
- $CE = CB$ (sides of square $BCED$)
- $\angle ACE = \angle ACF + \angle FCE = 90^\circ + \angle FCE$
- $\angle FCB = \angle BCE + \angle FCE = 90^\circ + \angle FCE$
- Therefore,$\angle ACE = \angle FCB$.
$4$. By $SAS$ congruence criterion,$\triangle ACE \cong \triangle FCB$. Since congruent triangles have equal areas,$\operatorname{ar}(\triangle ACE) = \operatorname{ar}(\triangle FCB)$.
$5$. Substituting this into the first equation: $\operatorname{ar}(CYXE) = 2 \operatorname{ar}(\triangle FCB)$.

(A) To prove $\operatorname{ar}(CYXE) = \operatorname{ar}(ACFG)$:
$1$. Consider $\Delta ABD$ and $\Delta MBC$. We have $AB = MB$ (sides of square $ABMN$),$BD = BC$ (sides of square $BCED$),and $\angle ABD = \angle ABC + \angle CBD = \angle ABC + 90^\circ = \angle ABC + \angle MBA = \angle MBC$.
$2$. By $SAS$ congruence,$\Delta ABD \cong \Delta MBC$.
$3$. Since $\Delta ABD$ and square $ABMN$ are on the same base $AB$ and between the same parallels,$\operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(ABMN)$.
$4$. Similarly,$\Delta MBC$ and rectangle $BYXD$ are on the same base $BD$ and between the same parallels,so $\operatorname{ar}(\Delta MBC) = \frac{1}{2} \operatorname{ar}(BYXD)$.
$5$. Since $\Delta ABD \cong \Delta MBC$,their areas are equal. Thus,$\frac{1}{2} \operatorname{ar}(ABMN) = \frac{1}{2} \operatorname{ar}(BYXD)$,which implies $\operatorname{ar}(ABMN) = \operatorname{ar}(BYXD)$.
$6$. By a similar argument,$\operatorname{ar}(ACFG) = \operatorname{ar}(CYXE)$.
$7$. Thus,$\operatorname{ar}(CYXE) = \operatorname{ar}(ACFG)$.

(N/A) To prove: $\operatorname{ar}(BCED) = \operatorname{ar}(ABMN) + \operatorname{ar}(ACFG)$.
$1$. Join $AD$ and $FC$. Observe $\triangle ABD$ and $\triangle MBC$. We have $AB = MB$ (sides of square $ABMN$),$BD = BC$ (sides of square $BCED$),and $\angle ABD = \angle ABC + 90^\circ = \angle MBC + 90^\circ = \angle MBC$. Thus,$\triangle ABD \cong \triangle MBC$ by $SAS$ congruence.
$2$. Since $\triangle ABD$ and square $ABMN$ are on the same base $AB$ and between the same parallels $AB$ and $MD$,$\operatorname{ar}(ABD) = \frac{1}{2} \operatorname{ar}(ABMN)$.
$3$. Similarly,$\triangle MBC$ and rectangle $BYXD$ are on the same base $BD$ and between the same parallels $BD$ and $CX$,so $\operatorname{ar}(MBC) = \frac{1}{2} \operatorname{ar}(BYXD)$.
$4$. Since $\triangle ABD \cong \triangle MBC$,their areas are equal. Therefore,$\frac{1}{2} \operatorname{ar}(ABMN) = \frac{1}{2} \operatorname{ar}(BYXD)$,which implies $\operatorname{ar}(ABMN) = \operatorname{ar}(BYXD)$.
$5$. By a similar argument,$\operatorname{ar}(ACFG) = \operatorname{ar}(CYXE)$.
$6$. Adding these,$\operatorname{ar}(ABMN) + \operatorname{ar}(ACFG) = \operatorname{ar}(BYXD) + \operatorname{ar}(CYXE) = \operatorname{ar}(BCED)$.