Textbook - Quadrilaterals Questions in English

(N/A) Let us recall what a rectangle is.
$A$ rectangle is a parallelogram in which one angle is a right angle.
Let $ABCD$ be a rectangle in which $\angle A = 90^{\circ}$.
We have to show that $\angle B = \angle C = \angle D = 90^{\circ}$.
We have $AD \parallel BC$ and $AB$ is a transversal.
So,$\angle A + \angle B = 180^{\circ}$ (Interior angles on the same side of the transversal are supplementary).
But,$\angle A = 90^{\circ}$.
So,$\angle B = 180^{\circ} - \angle A = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Now,$\angle C = \angle A$ and $\angle D = \angle B$ (Opposite angles of a parallelogram are equal).
So,$\angle C = 90^{\circ}$ and $\angle D = 90^{\circ}$.
Therefore,each of the angles of a rectangle is a right angle.

(N/A) Consider the rhombus $ABCD$ (see Fig.).
You know that $AB = BC = CD = DA$ (by definition of a rhombus).
Now,in $\Delta AOD$ and $\Delta COD$,
$OA = OC$ (Diagonals of a parallelogram bisect each other).
$OD = OD$ (Common side).
$AD = CD$ (Sides of a rhombus are equal).
Therefore,$\Delta AOD \cong \Delta COD$ ($SSS$ congruence rule).
This gives,$\angle AOD = \angle COD$ $(CPCT)$.
But,$\angle AOD + \angle COD = 180^{\circ}$ (Linear pair).
So,$2 \angle AOD = 180^{\circ}$.
Or,$\angle AOD = 90^{\circ}$.
Thus,the diagonals of a rhombus are perpendicular to each other.

(N/A) In $\Delta ABC$,we have $AB = AC$ (Given).
So,$\angle ABC = \angle ACB$ (Angles opposite to equal sides).
Also,$\angle PAC = \angle ABC + \angle ACB$ (Exterior angle property of a triangle).
Since $\angle ABC = \angle ACB$,we can write:
$\angle PAC = \angle ACB + \angle ACB = 2 \angle ACB$ ........... $(1)$
Now,$AD$ bisects $\angle PAC$,which means:
$\angle PAC = 2 \angle DAC$ ........... $(2)$
From equations $(1)$ and $(2)$,we get:
$2 \angle DAC = 2 \angle ACB$
Therefore,$\angle DAC = \angle ACB$ or $\angle DAC = \angle BCA$.

(N/A) Given: $AB = AC$ in $\triangle ABC$. Therefore,$\angle ABC = \angle ACB$.
Since $AD$ is the bisector of exterior angle $PAC$,we have $\angle PAD = \angle CAD = \frac{1}{2} \angle PAC$.
We know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles. Thus,$\angle PAC = \angle ABC + \angle ACB$.
Since $\angle ABC = \angle ACB$,we have $\angle PAC = 2 \angle ACB$.
Substituting this into the bisector equation: $\angle CAD = \frac{1}{2} (2 \angle ACB) = \angle ACB$.
These are alternate interior angles for lines $BC$ and $AD$ with transversal $AC$. Since the alternate interior angles are equal,$BC \parallel AD$.
We are given that $CD \parallel AB$.
Since both pairs of opposite sides are parallel ($BC \parallel AD$ and $AB \parallel CD$),$ABCD$ is a parallelogram.

(N/A) It is given that $l \parallel m$ and transversal $p$ intersects them at points $A$ and $C$ respectively.
The bisectors of $\angle PAC$ and $\angle ACQ$ intersect at $B$,and the bisectors of $\angle SAC$ and $\angle ACR$ intersect at $D$.
We need to show that quadrilateral $ABCD$ is a rectangle.
Since $l \parallel m$ and $p$ is a transversal,$\angle PAC = \angle ACR$ (Alternate interior angles).
Therefore,$\frac{1}{2} \angle PAC = \frac{1}{2} \angle ACR$,which implies $\angle BAC = \angle ACD$.
These are alternate interior angles for lines $AB$ and $DC$ with $AC$ as a transversal,and since they are equal,$AB \parallel DC$.
Similarly,by considering $\angle ACB$ and $\angle CAD$,we can show that $BC \parallel AD$.
Since both pairs of opposite sides are parallel,$ABCD$ is a parallelogram.
Now,$\angle PAC + \angle CAS = 180^{\circ}$ (Linear pair).
Dividing by $2$,we get $\frac{1}{2} \angle PAC + \frac{1}{2} \angle CAS = 90^{\circ}$,which means $\angle BAC + \angle CAD = 90^{\circ}$,so $\angle BAD = 90^{\circ}$.
Since $ABCD$ is a parallelogram with one angle equal to $90^{\circ}$,it is a rectangle.

(N/A) Let $P, Q, R$ and $S$ be the points of intersection of the bisectors of $\angle A$ and $\angle B$,$\angle B$ and $\angle C$,$\angle C$ and $\angle D$,and $\angle D$ and $\angle A$ respectively of parallelogram $ABCD$ (see figure).
In $\Delta ASD$,since $DS$ bisects $\angle D$ and $AS$ bisects $\angle A$,we have:
$\angle DAS + \angle ADS = \frac{1}{2} \angle A + \frac{1}{2} \angle D = \frac{1}{2} (\angle A + \angle D)$
Since $\angle A$ and $\angle D$ are interior angles on the same side of the transversal,$\angle A + \angle D = 180^{\circ}$.
Therefore,$\angle DAS + \angle ADS = \frac{1}{2} \times 180^{\circ} = 90^{\circ}$.
Using the angle sum property of a triangle in $\Delta ASD$:
$\angle DAS + \angle ADS + \angle DSA = 180^{\circ}$
$90^{\circ} + \angle DSA = 180^{\circ} \implies \angle DSA = 90^{\circ}$.
Since $\angle PSR$ and $\angle DSA$ are vertically opposite angles,$\angle PSR = 90^{\circ}$.
Similarly,it can be shown that $\angle SPQ = 90^{\circ}$,$\angle PQR = 90^{\circ}$,and $\angle SRQ = 90^{\circ}$.
Since all angles of the quadrilateral $PQRS$ are $90^{\circ}$,$PQRS$ is a rectangle.

(N/A) In quadrilateral $APCQ,$
$AP \parallel QC$ (since $AB \parallel CD$ and $P, Q$ are mid-points of $AB$ and $CD$ respectively) ............ $(1)$
$AP = \frac{1}{2} AB$ and $CQ = \frac{1}{2} CD$ (Given that $P$ and $Q$ are mid-points)
Also,$AB = CD$ (Opposite sides of a parallelogram are equal)
Since $AB = CD$,then $\frac{1}{2} AB = \frac{1}{2} CD$,which implies $AP = QC$ ............ $(2)$
From $(1)$ and $(2)$,since one pair of opposite sides is equal and parallel,$APCQ$ is a parallelogram.

(N/A) Given that $ABCD$ is a parallelogram,so $AB \,|| \,DC$ and $AB = DC$.
Since $P$ and $Q$ are mid-points of $AB$ and $CD$ respectively,we have $PB = \frac{1}{2} AB$ and $DQ = \frac{1}{2} DC$.
Since $AB = DC$,it follows that $\frac{1}{2} AB = \frac{1}{2} DC$,which implies $PB = DQ$.
Also,since $AB \,|| \,DC$,we have $PB \,|| \,DQ$.
In quadrilateral $DPBQ$,we have one pair of opposite sides ($PB$ and $DQ$) equal and parallel.
Therefore,$DPBQ$ is a parallelogram.

(N/A) Given: $ABCD$ is a parallelogram,$P$ is the mid-point of $AB$,and $Q$ is the mid-point of $CD$.
Since $AB \,||\, CD$ and $AB = CD$,we have $AP \,||\, QC$ and $AP = \frac{1}{2} AB = \frac{1}{2} CD = QC$.
Since $AP \,||\, QC$ and $AP = QC$,$APCQ$ is a parallelogram.
Therefore,$AQ \,||\, PC$,which implies $SQ \,||\, PR$.
Similarly,$PB \,||\, DQ$ and $PB = \frac{1}{2} AB = \frac{1}{2} CD = DQ$.
Since $PB \,||\, DQ$ and $PB = DQ$,$PBQD$ is a parallelogram.
Therefore,$DP \,||\, BQ$,which implies $SP \,||\, QR$.
Since both pairs of opposite sides are parallel ($SQ \,||\, PR$ and $SP \,||\, QR$),$PSQR$ is a parallelogram.

(A) Let the angles of the quadrilateral be $3x, 5x, 9x$,and $13x$.
Since the sum of all the interior angles of a quadrilateral is $360^{\circ}$,we have:
$3x + 5x + 9x + 13x = 360^{\circ}$
Combining the terms:
$30x = 360^{\circ}$
Solving for $x$:
$x = \frac{360^{\circ}}{30} = 12^{\circ}$
Now,calculate each angle:
$3x = 3 \times 12^{\circ} = 36^{\circ}$
$5x = 5 \times 12^{\circ} = 60^{\circ}$
$9x = 9 \times 12^{\circ} = 108^{\circ}$
$13x = 13 \times 12^{\circ} = 156^{\circ}$
Thus,the angles of the quadrilateral are $36^{\circ}, 60^{\circ}, 108^{\circ}$,and $156^{\circ}$.

(N/A) Let $ABCD$ be a parallelogram such that its diagonals are equal,i.e.,$AC = BD$.
In $\Delta ABC$ and $\Delta BAD$:
$AC = BD$ [Given]
$BC = AD$ [Opposite sides of a parallelogram are equal]
$AB = AB$ [Common side]
Therefore,by $SSS$ congruence criterion,$\Delta ABC \cong \Delta BAD$.
By $CPCT$,we have $\angle ABC = \angle BAD$ ............. $(1)$
Since $ABCD$ is a parallelogram,$AD \parallel BC$ and $AB$ is a transversal. Therefore,the sum of consecutive interior angles is $180^{\circ}$:
$\angle ABC + \angle BAD = 180^{\circ}$ [Consecutive interior angles are supplementary] ............. $(2)$
From $(1)$ and $(2)$,we get:
$\angle ABC + \angle ABC = 180^{\circ}$
$2 \angle ABC = 180^{\circ}$
$\angle ABC = 90^{\circ}$
Since $ABCD$ is a parallelogram with one angle equal to $90^{\circ}$,it is a rectangle.

(N/A) Let $ABCD$ be a quadrilateral such that its diagonals $AC$ and $BD$ bisect each other at right angles at point $O$.
In $\Delta AOB$ and $\Delta AOD$:
$AO = AO$ (Common side)
$OB = OD$ (Given that $O$ is the midpoint of $BD$)
$\angle AOB = \angle AOD = 90^{\circ}$ (Given)
By $SAS$ congruence criterion,$\Delta AOB \cong \Delta AOD$.
Therefore,$AB = AD$ (Corresponding parts of congruent triangles are equal) ... $(1)$
Similarly,by considering other pairs of triangles:
In $\Delta AOB$ and $\Delta COB$,we get $AB = CB$ ... $(2)$
In $\Delta COB$ and $\Delta COD$,we get $CB = CD$ ... $(3)$
In $\Delta COD$ and $\Delta AOD$,we get $CD = AD$ ... $(4)$
From $(1), (2), (3),$ and $(4)$,we have $AB = BC = CD = DA$.
Since all sides of the quadrilateral $ABCD$ are equal,it is a rhombus.

(N/A) We have a square $ABCD$ such that its diagonals $AC$ and $BD$ intersect at $O$.
$(i)$ To prove that the diagonals are equal,i.e.,$AC = BD$:
In $\Delta ABC$ and $\Delta BAD$,we have:
$AB = BA$ [Common]
$BC = AD$ [Opposite sides of the square $ABCD$]
$\angle ABC = \angle BAD = 90^{\circ}$ [All angles of a square are equal to $90^{\circ}$]
$\therefore \Delta ABC \cong \Delta BAD$ [$SAS$ criteria]
$\Rightarrow AC = BD$ ............. $(1)$
$(ii)$ To prove that $O$ is the mid-point of $AC$ and $BD$:
Because $AD \parallel BC$ and $AC$ is a transversal,$\angle 1 = \angle 3$ [Interior alternate angles].
Similarly,$\angle 2 = \angle 4$ [Interior alternate angles].
In $\Delta OAD$ and $\Delta OCB$,we have:
$AD = CB$ [Opposite sides of the square $ABCD$]
$\angle 1 = \angle 3$ [Proved]
$\angle 2 = \angle 4$ [Proved]
$\therefore \Delta OAD \cong \Delta OCB$ [$ASA$ criteria]
$\Rightarrow OA = OC$ and $OD = OB$
$\Rightarrow O$ is the mid-point of $AC$ and $BD$,i.e.,the diagonals $AC$ and $BD$ bisect each other at $O$. .......... $(2)$
$(iii)$ To prove that $AC \perp BD$:
In $\Delta OBA$ and $\Delta ODA$,we have:
$OB = OD$ [Proved]
$BA = DA$ [Opposite sides of the square]
$OA = OA$ [Common]
$\therefore \Delta OBA \cong \Delta ODA$ [$SSS$ criteria]
$\Rightarrow \angle AOB = \angle AOD$
Since $\angle AOB$ and $\angle AOD$ form a linear pair,$\angle AOB + \angle AOD = 180^{\circ}$.
$\Rightarrow \angle AOB = \angle AOD = 90^{\circ}$
$\Rightarrow AC \perp BD$ ............. $(3)$
From $(1)$,$(2)$,and $(3)$,we conclude that the diagonals of a square are equal and bisect each other at right angles.

(N/A) Let $ABCD$ be a quadrilateral such that its diagonals $AC$ and $BD$ are equal $(AC = BD)$ and bisect each other at right angles at point $O$ ($AO = OC$,$BO = OD$,and $\angle AOB = \angle BOC = \angle COD = \angle DOA = 90^{\circ}$).
$1$. To prove $ABCD$ is a rhombus:
In $\Delta AOD$ and $\Delta AOB$:
$AO = AO$ (Common)
$OD = OB$ (Given)
$\angle AOD = \angle AOB = 90^{\circ}$ (Given)
By $SAS$ congruence criterion,$\Delta AOD \cong \Delta AOB$.
Therefore,$AD = AB$ $(CPCT)$.
Similarly,we can prove $AB = BC$,$BC = CD$,and $CD = DA$.
Thus,$AB = BC = CD = DA$. Since all sides are equal,$ABCD$ is a rhombus.
$2$. To prove $ABCD$ is a square:
In $\Delta ABC$ and $\Delta BAD$:
$AC = BD$ (Given)
$BC = AD$ (Sides of rhombus are equal)
$AB = BA$ (Common)
By $SSS$ congruence criterion,$\Delta ABC \cong \Delta BAD$.
Therefore,$\angle ABC = \angle BAD$ $(CPCT)$.
Since $AD \parallel BC$ and $AB$ is a transversal,the sum of consecutive interior angles is $180^{\circ}$:
$\angle ABC + \angle BAD = 180^{\circ}$.
Since $\angle ABC = \angle BAD$,we have $2 \angle ABC = 180^{\circ}$,which implies $\angle ABC = 90^{\circ}$.
$A$ rhombus with one angle equal to $90^{\circ}$ is a square. Hence,$ABCD$ is a square.

(N/A) Given: $ABCD$ is a parallelogram in which diagonal $AC$ bisects $\angle A$. Thus,$\angle DAC = \angle BAC$.
To prove: $AC$ bisects $\angle C$,i.e.,$\angle DCA = \angle BCA$.
Proof:
$1$. Since $ABCD$ is a parallelogram,$AB \parallel DC$ and $AC$ is a transversal.
Therefore,$\angle BAC = \angle DCA$ (Alternate interior angles) ....... $(1)$
$2$. Also,$BC \parallel AD$ and $AC$ is a transversal.
Therefore,$\angle DAC = \angle BCA$ (Alternate interior angles) ....... $(2)$
$3$. Since $AC$ bisects $\angle A$,we have $\angle DAC = \angle BAC$ ....... $(3)$
$4$. From equations $(1)$,$(2)$,and $(3)$,we get:
$\angle DCA = \angle BCA$
Hence,$AC$ bisects $\angle C$.

(N/A) Given: $A$ parallelogram $ABCD$ in which diagonal $AC$ bisects $\angle A$. So,$\angle DAC = \angle BAC$.
To prove: $ABCD$ is a rhombus.
Proof:
$1$. Since $ABCD$ is a parallelogram,$AD \parallel BC$ and $AC$ is a transversal.
Therefore,$\angle DAC = \angle BCA$ (Alternate interior angles).
$2$. We are given that $\angle DAC = \angle BAC$.
$3$. From steps $1$ and $2$,we get $\angle BAC = \angle BCA$.
$4$. In $\Delta ABC$,since $\angle BAC = \angle BCA$,the sides opposite to these angles must be equal.
Therefore,$BC = AB$ (Sides opposite to equal angles are equal).
$5$. In a parallelogram,opposite sides are equal,so $AB = CD$ and $AD = BC$.
$6$. Since $AB = BC$ and $AB = CD, BC = AD$,we have $AB = BC = CD = DA$.
Since all sides of the parallelogram $ABCD$ are equal,it is a rhombus.

(N/A) $ABCD$ is a rhombus.
$AB = BC = CD = AD$
Also,$AB \parallel CD$ and $AD \parallel BC$.
Now,in $\triangle ADC$,$AD = CD$ (sides of a rhombus).
Therefore,$\angle 1 = \angle 2$ (angles opposite to equal sides are equal) ....... $(1)$
Also,$CD \parallel AB$ (opposite sides of a parallelogram) and $AC$ is a transversal.
Therefore,$\angle 2 = \angle 3$ (alternate interior angles) ....... $(2)$
From $(1)$ and $(2)$,we have $\angle 1 = \angle 3$.
Similarly,since $AD \parallel BC$ and $AC$ is a transversal,$\angle 1 = \angle 4$ (alternate interior angles).
Since $\angle 1 = \angle 2$ and $\angle 1 = \angle 4$,we have $\angle 2 = \angle 4$ is not necessarily true,but rather $\angle 1 = \angle 4$ and $\angle 2 = \angle 3$ implies $AC$ bisects $\angle A$ and $\angle C$.
Thus,$AC$ bisects $\angle A$ and $\angle C$.
Similarly,we can prove that $BD$ bisects $\angle B$ and $\angle D$.

(N/A) We have a rectangle $ABCD$ such that $AC$ bisects $\angle A$ as well as $\angle C$.
i.e.,$\angle 1 = \angle 4$ and $\angle 2 = \angle 3$ ....... $(1)$
Since a rectangle is a parallelogram,$ABCD$ is a parallelogram.
$\Rightarrow AB \parallel CD$ and $AC$ is a transversal.
$\therefore \angle 2 = \angle 4$ $[\text{Alternate interior angles}]$ .......... $(2)$
From $(1)$ and $(2)$,we have $\angle 3 = \angle 4$.
$\Rightarrow AB = BC$ $[\because \text{sides opposite to equal angles in } \Delta ABC \text{ are equal}]$.
Since $ABCD$ is a rectangle,$AB = CD$ and $BC = AD$.
$\therefore AB = BC = CD = AD$.
Thus,$ABCD$ is a rectangle having all of its sides equal,which means $ABCD$ is a square.

(N/A) Given: $ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ and $\angle C$.
$1$. Since $ABCD$ is a rectangle,$AB \parallel DC$ and $AD \parallel BC$.
$2$. Since $AC$ bisects $\angle A$,$\angle DAC = \angle BAC$. Also,$\angle DAC = \angle BCA$ (alternate interior angles).
$3$. Therefore,$\angle BAC = \angle BCA$. In $\triangle ABC$,sides opposite to equal angles are equal,so $AB = BC$.
$4$. $A$ rectangle with adjacent sides equal is a square. Thus,$ABCD$ is a square.
$5$. In a square,diagonals bisect the opposite angles.
$6$. Therefore,diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

(N/A) We have parallelogram $ABCD$. $BD$ is a diagonal and $P$ and $Q$ are points on $BD$ such that:
$DP = BQ$ [Given]
To prove that $\Delta APD \cong \Delta CQB$:
Since $AD \parallel BC$ and $BD$ is a transversal,$[\because ABCD$ is a parallelogram $]$
$\therefore \angle ADB = \angle CBD$ [Interior alternate angles]
$\Rightarrow \angle ADP = \angle CBQ$
Now,in $\Delta APD$ and $\Delta CQB$,we have:
$AD = CB$ [Opposite sides of a parallelogram are equal]
$DP = BQ$ [Given]
$\angle ADP = \angle CBQ$ [Proved above]
$\therefore$ By $SAS$ congruence criterion,we have:
$\Delta APD \cong \Delta CQB$.

(N/A) Given: $ABCD$ is a parallelogram. $P$ and $Q$ are points on diagonal $BD$ such that $DP = BQ$.
To prove: $AP = CQ$.
Proof:
In $\Delta APD$ and $\Delta CQB$:
$1$. $AD = CB$ (Opposite sides of a parallelogram are equal)
$2$. $\angle ADP = \angle CBQ$ (Alternate interior angles as $AD \parallel BC$ and $BD$ is a transversal)
$3$. $DP = BQ$ (Given)
Therefore,by $SAS$ congruence rule,$\Delta APD \cong \Delta CQB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AP = CQ$.

(N/A) We have parallelogram $ABCD$. $BD$ is a diagonal and $P$ and $Q$ are such that:
$PD = QB$ [Given]
To prove that $\Delta AQB \cong \Delta CPD$.
Since $AB \parallel CD$ and $BD$ is a transversal,$[\because ABCD$ is a parallelogram$]$
$\therefore \angle ABD = \angle CDB$
$\Rightarrow \angle ABQ = \angle CDP$
Now,in $\Delta AQB$ and $\Delta CPD$,we have:
$QB = PD$ [Given]
$\angle ABQ = \angle CDP$ [Proved above]
$AB = CD$ [Opposite sides of parallelogram $ABCD$]
$\therefore \Delta AQB \cong \Delta CPD$ [By $SAS$ congruence criterion]

(N/A) Given: $ABCD$ is a parallelogram. $P$ and $Q$ are points on diagonal $BD$ such that $DP = BQ$.
To prove: $AQ = CP$.
Proof:
In $\Delta AQB$ and $\Delta CPD$:
$1$. $AB = CD$ (Opposite sides of a parallelogram are equal)
$2$. $\angle ABQ = \angle CDP$ (Alternate interior angles as $AB \parallel CD$ and $BD$ is a transversal)
$3$. $BQ = DP$ (Given)
Therefore,by $SAS$ congruence criterion,$\Delta AQB \cong \Delta CPD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AQ = CP$.

(N/A) We have parallelogram $ABCD$. $BD$ is a diagonal and points $P$ and $Q$ are such that $DP = BQ$ [Given].
To prove that $APCQ$ is a parallelogram.
Let us join $AC$ intersecting $BD$ at $O$.
Since the diagonals of a parallelogram bisect each other,we have $AO = CO$ and $BO = DO$ ... $(1)$.
We are given $DP = BQ$ ... $(2)$.
Subtracting equation $(2)$ from $BO = DO$,we get:
$BO - BQ = DO - DP$
$QO = PO$ ... $(3)$.
Now,in quadrilateral $APCQ$,the diagonals $AC$ and $PQ$ bisect each other at $O$ (from equations $(1)$ and $(3)$).
Since the diagonals of a quadrilateral bisect each other,$APCQ$ is a parallelogram.

(N/A) In $\Delta APB$ and $\Delta CQD$,we have:
$1$. $\angle APB = \angle CQD = 90^{\circ}$ (Given,as $AP \perp BD$ and $CQ \perp BD$)
$2$. $AB = CD$ (Opposite sides of a parallelogram are equal)
$3$. $\angle ABP = \angle CDQ$ (Alternate interior angles,since $AB \parallel CD$ and $BD$ is a transversal)
Therefore,by the $AAS$ (Angle-Angle-Side) congruence criterion,we have:
$\Delta APB \cong \Delta CQD$

(N/A) In $\Delta APB$ and $\Delta CQD$:
$1$. $\angle APB = \angle CQD = 90^\circ$ (Given that $AP \perp BD$ and $CQ \perp BD$)
$2$. $AB = CD$ (Opposite sides of a parallelogram are equal)
$3$. $\angle ABP = \angle CDQ$ (Alternate interior angles,as $AB \parallel CD$ and $BD$ is a transversal)
Therefore,by the $AAS$ congruence criterion,$\Delta APB \cong \Delta CQD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AP = CQ$.

(N/A) To prove that $ABED$ is a parallelogram.
We know that a quadrilateral is a parallelogram if a pair of opposite sides is parallel and equal in length.
Given:
$AB = DE$
$AB \parallel DE$
In quadrilateral $ABED$,we have a pair of opposite sides ($AB$ and $DE$) which are both parallel and equal in length.
Therefore,$ABED$ is a parallelogram.

(N/A) To prove that $BEFC$ is a parallelogram.
Given that $BC = EF$ and $BC \parallel EF$.
$A$ quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel.
Since $BEFC$ is a quadrilateral in which the pair of opposite sides $BC$ and $EF$ is equal and parallel,
Therefore,$BEFC$ is a parallelogram.

(N/A) To prove that $AD \parallel CF$ and $AD = CF$:
$1$. In quadrilateral $ABED$,we are given that $AB = DE$ and $AB \parallel DE$. Since one pair of opposite sides is equal and parallel,$ABED$ is a parallelogram.
Therefore,$AD \parallel BE$ and $AD = BE$ (Opposite sides of a parallelogram are equal and parallel) ... $(1)$
$2$. In quadrilateral $BEFC$,we are given that $BC = EF$ and $BC \parallel EF$. Since one pair of opposite sides is equal and parallel,$BEFC$ is a parallelogram.
Therefore,$BE \parallel CF$ and $BE = CF$ (Opposite sides of a parallelogram are equal and parallel) ... $(2)$
$3$. From equations $(1)$ and $(2)$,we have $AD \parallel BE$ and $BE \parallel CF$,which implies $AD \parallel CF$.
Also,$AD = BE$ and $BE = CF$,which implies $AD = CF$.

(N/A) Given: In $\Delta ABC$ and $\Delta DEF$,$AB = DE$,$AB \parallel DE$,$BC = EF$ and $BC \parallel EF$.
Step $1$: Consider quadrilateral $ABED$.
Since $AB = DE$ and $AB \parallel DE$,one pair of opposite sides is equal and parallel.
Therefore,$ABED$ is a parallelogram.
This implies $AD = BE$ and $AD \parallel BE$ (Opposite sides of a parallelogram are equal and parallel).
Step $2$: Consider quadrilateral $BCFE$.
Since $BC = EF$ and $BC \parallel EF$,one pair of opposite sides is equal and parallel.
Therefore,$BCFE$ is a parallelogram.
This implies $BE = CF$ and $BE \parallel CF$ (Opposite sides of a parallelogram are equal and parallel).
Step $3$: Consider quadrilateral $ACFD$.
From Step $1$,$AD \parallel BE$ and from Step $2$,$BE \parallel CF$. Thus,$AD \parallel CF$.
From Step $1$,$AD = BE$ and from Step $2$,$BE = CF$. Thus,$AD = CF$.
Since one pair of opposite sides ($AD$ and $CF$) is equal and parallel,quadrilateral $ACFD$ is a parallelogram.

(N/A) Given: $AB = DE$,$AB \parallel DE$,$BC = EF$,and $BC \parallel EF$.
Step $1$: Consider quadrilateral $ABED$. Since $AB = DE$ and $AB \parallel DE$,one pair of opposite sides is equal and parallel. Therefore,$ABED$ is a parallelogram.
Step $2$: Since $ABED$ is a parallelogram,$AD = BE$ and $AD \parallel BE$.
Step $3$: Consider quadrilateral $BCFE$. Since $BC = EF$ and $BC \parallel EF$,one pair of opposite sides is equal and parallel. Therefore,$BCFE$ is a parallelogram.
Step $4$: Since $BCFE$ is a parallelogram,$BE = CF$ and $BE \parallel CF$.
Step $5$: From Step $2$ and Step $4$,we have $AD = BE$ and $BE = CF$,which implies $AD = CF$. Also,$AD \parallel BE$ and $BE \parallel CF$,which implies $AD \parallel CF$.
Step $6$: In quadrilateral $ACFD$,since $AD = CF$ and $AD \parallel CF$,it is a parallelogram.
Step $7$: Since $ACFD$ is a parallelogram,its opposite sides are equal. Therefore,$AC = DF$.

(N/A) Given: $AB = DE$,$AB \parallel DE$,$BC = EF$,and $BC \parallel EF$.
Step $1$: Since $AB = DE$ and $AB \parallel DE$,quadrilateral $ABED$ is a parallelogram. Therefore,$AD = BE$ and $AD \parallel BE$.
Step $2$: Since $BC = EF$ and $BC \parallel EF$,quadrilateral $BEFC$ is a parallelogram. Therefore,$BE = CF$ and $BE \parallel CF$.
Step $3$: From Step $1$ and Step $2$,we have $AD = BE$ and $BE = CF$,which implies $AD = CF$. Also,$AD \parallel CF$ because both are parallel to $BE$.
Step $4$: Since $AD = CF$ and $AD \parallel CF$,quadrilateral $ACFD$ is a parallelogram. Therefore,$AC = DF$.
Step $5$: In $\Delta ABC$ and $\Delta DEF$:
$AB = DE$ (Given)
$BC = EF$ (Given)
$AC = DF$ (Proved in Step $4$)
Therefore,by $SSS$ congruence criterion,$\Delta ABC \cong \Delta DEF$.

(N/A) We have $AB \parallel CD$ and $AD = BC$.
To prove that $\angle A = \angle B$.
Produce $AB$ to $E$ and draw $CE \parallel AD$.
$\therefore AB \parallel DC \Rightarrow AE \parallel DC$ [Given].
Also $AD \parallel CE$ [Construction].
$\therefore AECD$ is a parallelogram.
$\Rightarrow AD = CE$ [Opposite sides of the parallelogram $AECD$].
But $AD = BC$ [Given].
$\therefore BC = CE$.
Now,in $\Delta BCE$,we have $BC = CE$.
$\Rightarrow \angle CBE = \angle CEB$ ......... $(1)$ [$\because$ Angles opposite to equal sides of a triangle are equal].
Also,$\angle ABC + \angle CBE = 180^{\circ}$ [Linear pair] ......... $(2)$.
And $\angle A + \angle CEB = 180^{\circ}$ [$\because$ Adjacent angles of a parallelogram are supplementary] ......... $(3)$.
From $(2)$ and $(3)$,we get $\angle ABC + \angle CBE = \angle A + \angle CEB$.
But $\angle CBE = \angle CEB$ [Using $(1)$].
$\therefore \angle ABC = \angle A$,or $\angle B = \angle A$.

(N/A) To prove that $\angle C = \angle D$.
Construction: Extend $AB$ to $E$ and draw a line through $C$ parallel to $AD$ intersecting $AB$ produced at $E$.
Since $AD \parallel CE$ and $AE \parallel DC$,$AECD$ is a parallelogram.
Therefore,$AD = CE$ (opposite sides of a parallelogram).
Given that $AD = BC$,it follows that $BC = CE$.
In $\triangle BCE$,since $BC = CE$,the angles opposite to these sides are equal,so $\angle CBE = \angle CEB$.
Also,$\angle ABC + \angle CBE = 180^{\circ}$ (linear pair).
Since $AECD$ is a parallelogram,$\angle A = \angle ADC$ and $\angle D + \angle A = 180^{\circ}$ (consecutive interior angles).
Also,$\angle CEB = \angle A$ (corresponding angles as $AD \parallel CE$).
Since $\angle D + \angle A = 180^{\circ}$ and $\angle C + \angle B = 180^{\circ}$,and using the properties of the parallelogram and isosceles triangle,we conclude $\angle C = \angle D$.

(N/A) To prove $\Delta ABC \cong \Delta BAD$:
Construction: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ produced at $E$.
$1$. Since $AD \parallel CE$ and $AE \parallel DC$,$AECD$ is a parallelogram.
$2$. Therefore,$AD = CE$ (opposite sides of a parallelogram).
$3$. Given $AD = BC$,so $BC = CE$. Thus,$\angle CEB = \angle CBE$ (angles opposite to equal sides in $\Delta BCE$).
$4$. Also,$\angle ABC + \angle CBE = 180^{\circ}$ (linear pair) and $\angle BAD + \angle ADC = 180^{\circ}$ (consecutive interior angles).
$5$. Since $AD \parallel CE$,$\angle ADC + \angle DCE = 180^{\circ}$.
$6$. By comparing these,we can establish $\angle ABC = \angle BAD$.
$7$. In $\Delta ABC$ and $\Delta BAD$:
- $AB = BA$ (Common side)
- $BC = AD$ (Given)
- $\angle ABC = \angle BAD$ (Proved above)
$8$. By $SAS$ congruence rule,$\Delta ABC \cong \Delta BAD$.

(A) Given: $ABCD$ is a trapezium with $AB \parallel CD$ and $AD = BC$.
Construction: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ produced at $E$.
Proof:
$1$. Since $AD \parallel CE$ (by construction) and $AE \parallel DC$ (given $AB \parallel DC$),$AECD$ is a parallelogram.
$2$. Therefore,$AD = CE$ (opposite sides of a parallelogram). Since $AD = BC$ (given),we have $BC = CE$.
$3$. In $\Delta BCE$,$BC = CE$,so $\angle CBE = \angle CEB$ (angles opposite to equal sides).
$4$. Also,$\angle ABC + \angle CBE = 180^\circ$ (linear pair). Since $\angle CEB = \angle CBE$,we have $\angle ABC + \angle CEB = 180^\circ$.
$5$. In $\Delta ABC$ and $\Delta BAD$:
- $AB = BA$ (common side)
- $AD = BC$ (given)
- $\angle DAB = \angle CBA$ (since $\angle D + \angle A = 180^\circ$ and $\angle C + \angle B = 180^\circ$,and $AD=BC$ implies an isosceles trapezium).
$6$. By $SAS$ congruence rule,$\Delta ABC \cong \Delta BAD$.
$7$. Therefore,$AC = BD$ (by $CPCT$).

(N/A) Since $D$ and $E$ are the mid-points of sides $AB$ and $BC$ of $\Delta ABC$,by the Mid-point Theorem,$DE \parallel AC$ and $DE = \frac{1}{2} AC = AF$.
Similarly,$DF \parallel BC$ and $EF \parallel AB$.
Therefore,$AFDE$,$BDFE$,and $DFCE$ are all parallelograms.
Now,$DE$ is a diagonal of the parallelogram $BDFE$,which divides it into two congruent triangles,so $\Delta BDE \cong \Delta FED$.
Similarly,$DF$ is a diagonal of the parallelogram $AFDE$,so $\Delta DAF \cong \Delta FED$.
Also,$EF$ is a diagonal of the parallelogram $DFCE$,so $\Delta EFC \cong \Delta FED$.
Since all three triangles are congruent to $\Delta FED$,it follows that all four triangles $(\Delta BDE, \Delta DAF, \Delta EFC, \text{ and } \Delta FED)$ are congruent to each other.

(N/A) We are given that $AB = BC$ and we have to prove that $DE = EF$.
Let us join $A$ to $F$ intersecting $m$ at $G$.
The quadrilateral $ADFC$ is divided into two triangles,namely $\Delta ACF$ and $\Delta AFD$.
In $\Delta ACF$,it is given that $B$ is the mid-point of $AC$ $(AB = BC)$ and $BG \parallel CF$ (since $m \parallel n$). By the converse of the mid-point theorem,$G$ is the mid-point of $AF$.
Now,in $\Delta AFD$,we have $G$ as the mid-point of $AF$ and $GE \parallel AD$ (since $m \parallel l$). By the converse of the mid-point theorem,$E$ is the mid-point of $DF$.
Therefore,$DE = EF$.
In other words,$l, m$ and $n$ cut off equal intercepts on $q$ also.

(N/A) Given: $ABCD$ is a quadrilateral,$P$,$Q$,$R$,$S$ are mid-points of sides $AB$,$BC$,$CD$,$DA$ respectively. $AC$ is a diagonal.
To prove: $SR \parallel AC$ and $SR = \frac{1}{2} AC$.
Proof: In $\Delta ACD$,we have:
$S$ is the mid-point of $AD$.
$R$ is the mid-point of $CD$.
According to the Mid-point Theorem,the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it.
Therefore,in $\Delta ACD$,the line segment $SR$ joins the mid-points of sides $AD$ and $CD$.
Hence,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.

(N/A) Given: $P, Q, R, S$ are mid-points of sides $AB, BC, CD, DA$ respectively of quadrilateral $ABCD$. $AC$ is a diagonal.
To prove: $PQ = SR$.
Proof:
In $\triangle ABC$,$P$ is the mid-point of $AB$ and $Q$ is the mid-point of $BC$.
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,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$ ........ $(1)$
In $\triangle ADC$,$S$ is the mid-point of $AD$ and $R$ is the mid-point of $CD$.
By the Mid-point Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$ ........ $(2)$
From equations $(1)$ and $(2)$,since both $PQ$ and $SR$ are equal to $\frac{1}{2} AC$,we get $PQ = SR$.

(N/A) To prove that $PQRS$ is a parallelogram.
In $\Delta ABC$,$P$ and $Q$ are the mid-points of $AB$ and $BC$.
$\therefore PQ = \frac{1}{2} AC$ and $PQ \parallel AC$ .......... $(1)$
In $\Delta ACD$,$S$ and $R$ are the mid-points of $DA$ and $CD$.
$\therefore SR = \frac{1}{2} AC$ and $SR \parallel AC$ .......... $(2)$
From $(1)$ and $(2)$,we get
$PQ = \frac{1}{2} AC = SR$ and $PQ \parallel AC \parallel SR$
$\Rightarrow PQ = SR$ and $PQ \parallel SR$
i.e.,one pair of opposite sides in quadrilateral $PQRS$ is equal and parallel.
$\therefore PQRS$ is a parallelogram.

(N/A) Given: $ABCD$ is a rhombus. $P, Q, R, S$ are mid-points of sides $AB, BC, CD, DA$ respectively.
To prove: $PQRS$ is a rectangle.
Construction: Join $AC$ and $BD$.
Proof:
$1$. In $\Delta ABC$,$P$ and $Q$ are mid-points of $AB$ and $BC$. By Mid-point Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$ ... $(1)$.
$2$. In $\Delta ADC$,$S$ and $R$ are mid-points of $AD$ and $CD$. By Mid-point Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$ ... $(2)$.
$3$. From $(1)$ and $(2)$,$PQ \parallel SR$ and $PQ = SR$. Since one pair of opposite sides is equal and parallel,$PQRS$ is a parallelogram.
$4$. In a rhombus,diagonals bisect each other at $90^{\circ}$. Let $AC$ and $BD$ intersect at $O$. Thus,$AC \perp BD$.
$5$. Since $PQ \parallel AC$ and $QR \parallel BD$,the angle between $PQ$ and $QR$ must be $90^{\circ}$ because the angle between $AC$ and $BD$ is $90^{\circ}$.
$6$. Since $PQRS$ is a parallelogram with one angle equal to $90^{\circ}$,$PQRS$ is a rectangle.

(N/A) In a rectangle $ABCD$,$P$ is the mid-point of $AB$,$Q$ is the mid-point of $BC$,$R$ is the mid-point of $CD$,and $S$ is the mid-point of $DA$.
Draw diagonal $AC$.
In $\Delta ABC$,by the Mid-point theorem:
$PQ = \frac{1}{2} AC$ and $PQ \parallel AC$ ......... $(1)$
In $\Delta ACD$,by the Mid-point theorem:
$SR = \frac{1}{2} AC$ and $SR \parallel AC$ ......... $(2)$
From $(1)$ and $(2)$,we get $PQ = SR$ and $PQ \parallel SR$.
Similarly,by joining $BD$,we have $PS = QR$ and $PS \parallel QR$.
Since both pairs of opposite sides of quadrilateral $PQRS$ are equal and parallel,$PQRS$ is a parallelogram.
Now,in $\Delta PAS$ and $\Delta PBQ$:
$\angle A = \angle B = 90^{\circ}$ (Angles of a rectangle)
$AP = BP$ ($P$ is the mid-point of $AB$)
$AS = BQ$ (Halves of equal opposite sides $AD$ and $BC$)
By $SAS$ congruence criterion,$\Delta PAS \cong \Delta PBQ$.
Therefore,$PS = PQ$ (Corresponding parts of congruent triangles).
Since $PQRS$ is a parallelogram with adjacent sides $PS = PQ$,it follows that all sides are equal $(PQ = QR = RS = SP)$.
Thus,$PQRS$ is a rhombus.

(N/A) In trapezium $ABCD$,$AB \parallel DC$. $E$ is the mid-point of $AD$. $EF$ is drawn parallel to $AB$. We have to prove that $F$ is the mid-point of $BC$.
Join $BD$.
In $\Delta DAB$:
Since $E$ is the mid-point of $AD$ [Given] and $EG \parallel AB$ [Since $EF \parallel AB$],by the converse of the mid-point theorem,$G$ is the mid-point of $BD$.
In $\Delta BDC$:
Since $G$ is the mid-point of $BD$ [Proved] and $GF \parallel DC$ [Since $AB \parallel DC$ and $EF \parallel AB$,therefore $GF \parallel DC$],by the converse of the mid-point theorem,$F$ is the mid-point of $BC$.

(N/A) Given: $ABCD$ is a parallelogram where $E$ and $F$ are mid-points of $AB$ and $CD$ respectively.
To prove: $AF$ and $EC$ trisect the diagonal $BD$,i.e.,$DP = PQ = QB$.
Proof:
Since $ABCD$ is a parallelogram,$AB \parallel DC$ and $AB = DC$.
Since $E$ and $F$ are mid-points,$AE = \frac{1}{2}AB$ and $FC = \frac{1}{2}DC$.
Since $AB = DC$,we have $AE = FC$.
Also,$AE \parallel FC$ because $AB \parallel DC$.
Thus,$AECF$ is a parallelogram (a quadrilateral with one pair of opposite sides equal and parallel).
Therefore,$AF \parallel EC$.
In $\Delta DQC$,$F$ is the mid-point of $DC$ and $FP \parallel CQ$ (since $AF \parallel EC$). By the converse of the mid-point theorem,$P$ is the mid-point of $DQ$,so $DP = PQ$ ... $(1)$.
In $\Delta ABP$,$E$ is the mid-point of $AB$ and $EQ \parallel AP$ (since $AF \parallel EC$). By the converse of the mid-point theorem,$Q$ is the mid-point of $BP$,so $PQ = QB$ ... $(2)$.
From $(1)$ and $(2)$,$DP = PQ = QB$.
Hence,$AF$ and $EC$ trisect the diagonal $BD$.

(N/A) Let $ABCD$ be a quadrilateral such that the mid-points of sides $AB$,$BC$,$CD$,and $DA$ are $P$,$Q$,$R$,and $S$ respectively. We need to prove that the line segments joining the mid-points of opposite sides,i.e.,$PR$ and $SQ$,bisect each other at $O$.
Join $PQ$,$QR$,$RS$,and $SP$. Also,join $AC$ and $BD$.
In $\Delta ABC$,$P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively.
By the Mid-point Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
In $\Delta ADC$,$S$ and $R$ are the mid-points of $AD$ and $CD$ respectively.
By the Mid-point Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.
From the above,$PQ \parallel SR$ and $PQ = SR$.
Since one pair of opposite sides of quadrilateral $PQRS$ is equal and parallel,$PQRS$ is a parallelogram.
We know that the diagonals of a parallelogram bisect each other.
Therefore,the diagonals $PR$ and $SQ$ of the parallelogram $PQRS$ bisect each other at $O$.
Hence,the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

(N/A) We have a triangle $ABC$ such that $\angle C = 90^{\circ}$. $M$ is the mid-point of $AB$ and $MD \parallel BC$.
To prove that $D$ is the mid-point of $AC$.
In $\Delta ACB$,we have:
$M$ as the mid-point of $AB$. [Given]
$MD \parallel BC$. [Given]
Therefore,using the converse of the mid-point theorem,$D$ is the mid-point of $AC$.

(N/A) We have a triangle $ABC$ such that $\angle C = 90^{\circ}$. $M$ is the mid-point of $AB$ and $MD \parallel BC$.
To prove: $MD \perp AC$.
Since $MD \parallel BC$ and $AC$ is a transversal,
$\therefore \angle MDA = \angle BCA$ [Corresponding angles].
But $\angle BCA = 90^{\circ}$ [Given].
$\therefore \angle MDA = 90^{\circ}$.
$\Rightarrow MD \perp AC$.

(N/A) Given: In $\Delta ABC$,$\angle C = 90^{\circ}$,$M$ is the mid-point of $AB$,and $MD \parallel BC$.
To prove: $CM = MA = \frac{1}{2} AB$.
Proof:
$1$. In $\Delta ABC$,since $MD \parallel BC$ and $M$ is the mid-point of $AB$,by the Converse of Mid-point Theorem,$D$ is the mid-point of $AC$. Thus,$AD = CD$.
$2$. Now,consider $\Delta ADM$ and $\Delta CDM$:
- $AD = CD$ (Proved above)
- $\angle ADM = \angle CDM = 90^{\circ}$ (Since $MD \parallel BC$ and $\angle ACB = 90^{\circ}$,$\angle ADM = \angle ACB = 90^{\circ}$ by corresponding angles)
- $DM = DM$ (Common side)
$3$. By $SAS$ congruence criterion,$\Delta ADM \cong \Delta CDM$.
$4$. By $c.p.c.t.$,$MA = MC$.
$5$. Since $M$ is the mid-point of $AB$,$MA = \frac{1}{2} AB$.
$6$. Therefore,$CM = MA = \frac{1}{2} AB$.