Mix Examples - Quadrilaterals Questions in English

(A) In a parallelogram $ABCD,$ the diagonals $AC$ and $BD$ intersect at $O.$
Given $\angle BOC = 90^{\circ}.$ Since $AC$ is a straight line,$\angle BOC + \angle AOB = 180^{\circ},$ so $\angle AOB = 90^{\circ}.$
In $\triangle BOC,$ the sum of angles is $180^{\circ}.$ Thus,$\angle OBC + \angle BOC + \angle OCB = 180^{\circ}.$
Since $AB \parallel DC,$ alternate interior angles are equal,so $\angle OAB = \angle OCD.$
Also,$\angle BDC = 50^{\circ}.$ Since $AB \parallel DC,$ $\angle ABD = \angle BDC = 50^{\circ}$ (alternate interior angles).
In $\triangle AOB,$ $\angle AOB = 90^{\circ}$ and $\angle ABO = 50^{\circ}.$
The sum of angles in $\triangle AOB$ is $180^{\circ},$
$\angle OAB + \angle AOB + \angle ABO = 180^{\circ}$
$\angle OAB + 90^{\circ} + 50^{\circ} = 180^{\circ}$
$\angle OAB + 140^{\circ} = 180^{\circ}$
$\angle OAB = 180^{\circ} - 140^{\circ} = 40^{\circ}.$

(B) The sum of all interior angles of a quadrilateral is $360^{\circ}$.
Let the fourth angle be $x$.
Then,$75^{\circ} + 90^{\circ} + 75^{\circ} + x = 360^{\circ}$.
$240^{\circ} + x = 360^{\circ}$.
$x = 360^{\circ} - 240^{\circ}$.
$x = 120^{\circ}$.

(C) Let the rectangle be $ABCD$ with diagonals $AC$ and $BD$ intersecting at $O$.
Given that the diagonal $AC$ is inclined to side $AB$ at an angle of $25^{\circ}$,so $\angle CAB = 25^{\circ}$.
In a rectangle,diagonals are equal and bisect each other,so $OA = OB$.
In $\triangle OAB$,since $OA = OB$,the angles opposite to these sides are equal,so $\angle OBA = \angle OAB = 25^{\circ}$.
Using the angle sum property in $\triangle OAB$:
$\angle AOB = 180^{\circ} - (25^{\circ} + 25^{\circ}) = 180^{\circ} - 50^{\circ} = 130^{\circ}$.
Since $\angle AOB$ and $\angle AOD$ form a linear pair:
$\angle AOD = 180^{\circ} - \angle AOB = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Thus,the acute angle between the diagonals is $50^{\circ}$.

(D) Given that $ABCD$ is a rhombus and $\angle ACB = 40^{\circ}$.
We know that the diagonals of a rhombus bisect each other at right angles $(90^{\circ})$.
Let the diagonals intersect at point $O$. Thus,$\triangle BOC$ is a right-angled triangle with $\angle BOC = 90^{\circ}$.
In $\triangle BOC$,the sum of angles is $180^{\circ}$:
$\angle OBC + \angle BOC + \angle BCO = 180^{\circ}$
$\angle OBC + 90^{\circ} + 40^{\circ} = 180^{\circ}$
$\angle OBC = 180^{\circ} - 130^{\circ} = 50^{\circ}$.
Since $AD \parallel BC$ (opposite sides of a rhombus are parallel),the alternate interior angles are equal:
$\angle ADB = \angle DBC$.
Since $\angle DBC = \angle OBC = 50^{\circ}$,we have $\angle ADB = 50^{\circ}$.

(A) Let the mid-points of sides $PQ, QR, RS,$ and $SP$ be $A, B, C,$ and $D$ respectively.
According to the Mid-point Theorem,in $\triangle PQS,$ $AD \parallel QS$ and $AD = \frac{1}{2} QS.$
Similarly,in $\triangle RQS,$ $BC \parallel QS$ and $BC = \frac{1}{2} QS.$
Thus,$AD \parallel BC$ and $AD = BC,$ which makes $ABCD$ a parallelogram.
For $ABCD$ to be a rectangle,its adjacent sides must be perpendicular (i.e.,$AB \perp AD$).
Since $AB \parallel PR$ and $AD \parallel QS,$ the condition $AB \perp AD$ implies $PR \perp QS.$
Therefore,the quadrilateral formed by joining the mid-points is a rectangle if the diagonals of the original quadrilateral $PQRS$ are perpendicular.

(B) Let the mid-points of sides $PQ, QR, RS,$ and $SP$ be $A, B, C,$ and $D$ respectively. By the Mid-point Theorem,$AB \parallel PR$ and $AB = \frac{1}{2} PR$. Similarly,$DC \parallel PR$ and $DC = \frac{1}{2} PR$. Thus,$AB \parallel DC$ and $AB = DC$,making $ABCD$ a parallelogram. For $ABCD$ to be a rhombus,adjacent sides must be equal (i.e.,$AB = AD$). Since $AD = \frac{1}{2} QS$ and $AB = \frac{1}{2} PR$,the condition $AB = AD$ implies $\frac{1}{2} PR = \frac{1}{2} QS$,which means $PR = QS$. Therefore,the quadrilateral formed by joining the mid-points is a rhombus if the diagonals of the original quadrilateral $PQRS$ are equal.

(C) Let the angles $A, B, C$ and $D$ of the quadrilateral $ABCD$ be $3x, 7x, 6x$ and $4x$ respectively.
Since the sum of the angles of a quadrilateral is $360^{\circ}$,we have:
$3x + 7x + 6x + 4x = 360^{\circ}$
$20x = 360^{\circ}$
$x = 18^{\circ}$
Thus,the angles are:
$A = 3 \times 18^{\circ} = 54^{\circ}$
$B = 7 \times 18^{\circ} = 126^{\circ}$
$C = 6 \times 18^{\circ} = 108^{\circ}$
$D = 4 \times 18^{\circ} = 72^{\circ}$
Now,consider the sum of consecutive angles $\angle C$ and $\angle D$:
$\angle C + \angle D = 108^{\circ} + 72^{\circ} = 180^{\circ}$
Since the sum of the interior angles on the same side of the transversal $CD$ is $180^{\circ}$,the sides $AD$ and $BC$ must be parallel $(AD \parallel BC)$.
Since $ABCD$ is a quadrilateral with one pair of opposite sides parallel,it is a trapezium.

(D) In quadrilateral $ABCD$,the sum of interior angles is $\angle A + \angle B + \angle C + \angle D = 360^{\circ}$.
In $\triangle APB$,$\angle P + \frac{1}{2}\angle A + \frac{1}{2}\angle B = 180^{\circ} \implies \angle P = 180^{\circ} - \frac{1}{2}(\angle A + \angle B)$.
Similarly,in $\triangle CRD$,$\angle R = 180^{\circ} - \frac{1}{2}(\angle C + \angle D)$.
Adding these,$\angle P + \angle R = 360^{\circ} - \frac{1}{2}(\angle A + \angle B + \angle C + \angle D) = 360^{\circ} - \frac{1}{2}(360^{\circ}) = 360^{\circ} - 180^{\circ} = 180^{\circ}$.
Since the sum of opposite angles $\angle P + \angle R = 180^{\circ}$,the quadrilateral $PQRS$ is a cyclic quadrilateral,meaning its opposite angles are supplementary.

(A) Let $PN$ be the bisector of $\angle APQ$ and $QN$ be the bisector of $\angle CQP$.
Since $APB \parallel CQD$,the sum of consecutive interior angles is $\angle APQ + \angle CQP = 180^\circ$.
Dividing by $2$,we get $\frac{1}{2}\angle APQ + \frac{1}{2}\angle CQP = 90^\circ$.
In $\triangle PNQ$,$\angle NPQ + \angle NQP = 90^\circ$,therefore $\angle PNQ = 90^\circ$.
Similarly,for the other vertices $M, P, Q$,we can show that all angles of the quadrilateral $PNQM$ are $90^\circ$.
Thus,the quadrilateral $PNQM$ formed by the angle bisectors is a rectangle.

(B) Let the rhombus be $ABCD$. Let $P, Q, R,$ and $S$ be the mid-points of sides $AB, BC, CD,$ and $DA$ respectively.
By the Mid-point Theorem,in $\triangle ABC$,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
Similarly,in $\triangle ADC$,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.
Thus,$PQ \parallel SR$ and $PQ = SR$,so $PQRS$ is a parallelogram.
Since the diagonals of a rhombus bisect each other at $90^\circ$,the sides of the quadrilateral $PQRS$ will be parallel to the diagonals $AC$ and $BD$.
Because the diagonals of a rhombus are perpendicular $(AC \perp BD)$,the adjacent sides of the quadrilateral $PQRS$ are perpendicular to each other.
$A$ parallelogram with one right angle is a rectangle. Therefore,$PQRS$ is a rectangle.

(C) In $\Delta ABC$,$D$ and $E$ are the mid-points of $AB$ and $AC$ respectively. By the Mid-point Theorem,$DE \parallel BC$ and $DE = \frac{1}{2} BC$.
In $\Delta ABO$,$D$ and $P$ are the mid-points of $AB$ and $OB$ respectively. By the Mid-point Theorem,$DP \parallel AO$ and $DP = \frac{1}{2} AO$.
In $\Delta ACO$,$E$ and $Q$ are the mid-points of $AC$ and $OC$ respectively. By the Mid-point Theorem,$EQ \parallel AO$ and $EQ = \frac{1}{2} AO$.
From the above,we have $DP = EQ$ (both are equal to $\frac{1}{2} AO$) and $DP \parallel EQ$ (both are parallel to $AO$).
Since one pair of opposite sides of the quadrilateral $DEQP$ is equal and parallel,$DEQP$ is a parallelogram.

(D) According to the Varignon's Theorem,the figure formed by joining the mid-points of the sides of any quadrilateral is a parallelogram.
For this parallelogram to be a square,the sides must be equal and the adjacent sides must be perpendicular.
The sides of the quadrilateral formed by joining the mid-points are parallel to the diagonals of the original quadrilateral $ABCD$ and are equal to half the length of the diagonals.
Therefore,for the resulting figure to be a square,the diagonals of the original quadrilateral $ABCD$ must be equal in length (to make the sides of the parallelogram equal) and perpendicular to each other (to make the angles of the parallelogram $90^{\circ}$).

(A) In parallelogram $ABCD,$ $AD \parallel BC$ and $AC$ is a transversal.
Therefore,$\angle DAC = \angle ACB$ (Alternate interior angles).
Given $\angle DAC = 32^{\circ},$ so $\angle ACB = 32^{\circ}.$
Now,consider $\Delta BOC.$ The angle $\angle AOB$ is an exterior angle to $\Delta BOC$ at vertex $O.$
By the exterior angle theorem,the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Therefore,$\angle AOB = \angle OCB + \angle OBC.$
Substituting the known values,$70^{\circ} = 32^{\circ} + \angle OBC.$
Thus,$\angle OBC = 70^{\circ} - 32^{\circ} = 38^{\circ}.$
Since $\angle DBC$ is the same as $\angle OBC,$ we have $\angle DBC = 38^{\circ}.$

(B) In a parallelogram,opposite sides are equal and parallel. Opposite angles are equal. Diagonals bisect each other. However,the diagonals do not necessarily bisect the opposite angles; this property is only true for a rhombus or a square. Therefore,the statement that 'opposite angles are bisected by the diagonals' is not true for a general parallelogram.

(C) To prove that $CF$ is equal and parallel to $DA$,we consider $\triangle ADE$ and $\triangle CFE$.
In these triangles:
$1$. $AE = CE$ (Since $E$ is the mid-point of $AC$)
$2$. $\angle AED = \angle CEF$ (Vertically opposite angles)
If we are given $DE = EF$,then by $SAS$ congruence criterion,$\triangle ADE \cong \triangle CFE$.
By $CPCT$,we get $AD = CF$ and $\angle ADE = \angle CFE$.
Since $\angle ADE$ and $\angle CFE$ are alternate interior angles,it implies $AD \parallel CF$.
Thus,the additional information required is $DE = EF$.

(C) parallelogram with equal diagonals is a rectangle.
In a rectangle,all interior angles are right angles.
Therefore,$\angle ABC = 90^{\circ}$.

(B) This statement is false.
In a rhombus,the diagonals are perpendicular bisectors of each other,but they are not necessarily equal in length.
Only in the case of a square (which is a special type of rhombus) are the diagonals equal.

(B) No,it is not necessarily a parallelogram. $A$ quadrilateral is a parallelogram only if both pairs of opposite angles are equal. For example,if $\angle A = \angle B = \angle C = 80^{\circ}$,then the sum of angles in a quadrilateral is $360^{\circ}$,so $\angle D = 360^{\circ} - (80^{\circ} + 80^{\circ} + 80^{\circ}) = 120^{\circ}$. Since $\angle B \neq \angle D$ $(80^{\circ} \neq 120^{\circ})$,the condition for a parallelogram is not satisfied.

(B) quadrilateral is a parallelogram if and only if its diagonals bisect each other,meaning $OA = OC$ and $OB = OD$.
In the given quadrilateral $ABCD$,the ratio of the segments of the diagonal $AC$ is $OA : OC = 3 : 2$.
Since $OA \neq OC$,the diagonals do not bisect each other.
Therefore,$ABCD$ is not a parallelogram.

(A) We know that the diagonals of a parallelogram bisect each other at their point of intersection.
Since $O$ is the intersection point,$AC = 2 \times OA$ and $BD = 2 \times OD$.
Given $OA = 3\, cm$ and $OD = 2\, cm$.
Therefore,$AC = 2 \times 3\, cm = 6\, cm$.
And $BD = 2 \times 2\, cm = 4\, cm$.
Hence,the lengths of $AC$ and $BD$ are $6\, cm$ and $4\, cm$ respectively.

(B) The statement is false.
In a parallelogram,the diagonals bisect each other,but they are not necessarily perpendicular.
Perpendicular diagonals are a property of a rhombus or a square,which are special types of parallelograms,but it is not a general property of all parallelograms.

(N/A) The sum of the given angles is $110^{\circ} + 80^{\circ} + 70^{\circ} + 95^{\circ} = 355^{\circ}$.
We know that the sum of the interior angles of a quadrilateral is always $360^{\circ}$.
Since the sum of the given angles $(355^{\circ})$ is not equal to $360^{\circ}$,these cannot be the angles of a quadrilateral.

(B) In quadrilateral $ABCD$,$\angle A + \angle D = 180^{\circ}$.
Since $\angle A$ and $\angle D$ are consecutive interior angles on the same side of the transversal $AD$ intersecting lines $AB$ and $CD$,their sum being $180^{\circ}$ implies that the lines $AB$ and $CD$ are parallel.
By definition,a quadrilateral with at least one pair of parallel opposite sides is called a trapezium.
Therefore,the special name that can be given to this quadrilateral $ABCD$ is a trapezium.

(B) The sum of all interior angles of a quadrilateral is $360^{\circ}$.
Let each angle of the quadrilateral be $x$.
Since all four angles are equal,we have $x + x + x + x = 360^{\circ}$.
$4x = 360^{\circ}$.
$x = 90^{\circ}$.
$A$ quadrilateral in which all angles are $90^{\circ}$ is called a rectangle.

(B) The given statement is false. While the diagonals of a rectangle are equal in length,they are not necessarily perpendicular to each other. The diagonals of a rectangle bisect each other,but they only intersect at $90^{\circ}$ if the rectangle is a square.

(N/A) No,all four angles of a quadrilateral cannot be obtuse angles.
An obtuse angle is an angle greater than $90^{\circ}$.
If all four angles were obtuse,each angle would be $> 90^{\circ}$.
Therefore,the sum of the four angles would be $> 90^{\circ} + 90^{\circ} + 90^{\circ} + 90^{\circ} = 360^{\circ}$.
However,the sum of the interior angles of a quadrilateral is always exactly $360^{\circ}$.
Since the sum cannot exceed $360^{\circ}$,it is impossible for all four angles to be obtuse.

(C) In $\triangle ABC$,$D$ is the mid-point of $AB$ and $E$ is the mid-point of $BC$.
According to 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 the length of the third side.
Therefore,$DE = \frac{1}{2} \times AC$.
Given $AC = 7\, cm$.
$DE = \frac{1}{2} \times 7\, cm = 3.5\, cm$.

(A) $BDEF$ is a parallelogram.
Therefore,$BD = EF$ ... $(1)$ [Opposite sides of a parallelogram are equal]
$FDCE$ is a parallelogram.
Therefore,$CD = EF$ ... $(2)$ [Opposite sides of a parallelogram are equal]
From equations $(1)$ and $(2)$,we get:
$BD = CD$
Yes,we can say that $BD = CD$ because both are equal to the same side $EF$.

(A) We know that opposite angles of a parallelogram are equal.
In parallelogram $ABCD$,we have $\angle A = \angle C$.
Given that $\angle C = 55^{\circ}$.
Therefore,$\angle A = 55^{\circ}$.
Now,in parallelogram $AEFG$,the opposite angles are equal,so $\angle F = \angle A$.
Since $\angle A = 55^{\circ}$,therefore $\angle F = 55^{\circ}$.

(N/A) No,all the angles of a quadrilateral cannot be acute angles.
An acute angle is defined as an angle less than $90^{\circ}$.
If all four angles of a quadrilateral were acute,each angle would be less than $90^{\circ}$.
Consequently,the sum of these four angles would be less than $90^{\circ} + 90^{\circ} + 90^{\circ} + 90^{\circ} = 360^{\circ}$.
However,the sum of the interior angles of a quadrilateral is always exactly $360^{\circ}$.
Therefore,it is impossible for all four angles to be acute.

(A) Yes,all the angles of a quadrilateral can be right angles.
According to the angle sum property of a quadrilateral,the sum of all interior angles is $360^{\circ}$.
If each angle is $90^{\circ}$,then the sum of the four angles is $90^{\circ} + 90^{\circ} + 90^{\circ} + 90^{\circ} = 360^{\circ}$.
Since this satisfies the angle sum property,a quadrilateral can have all right angles (e.g.,a rectangle or a square).

(A) Since the diagonals of a quadrilateral $ABCD$ bisect each other,$ABCD$ must be a parallelogram.
In a parallelogram,the sum of any two adjacent angles is $180^{\circ}$ because they are supplementary.
Therefore,$\angle A + \angle B = 180^{\circ}$.
Given that $\angle A = 35^{\circ}$,we substitute this value into the equation:
$35^{\circ} + \angle B = 180^{\circ}$.
Solving for $\angle B$:
$\angle B = 180^{\circ} - 35^{\circ} = 145^{\circ}$.

(B) quadrilateral in which both pairs of opposite angles are equal is a parallelogram.
Since the opposite angles of quadrilateral $ABCD$ are equal,$ABCD$ is a parallelogram.
In a parallelogram,opposite sides are equal in length.
Therefore,$AB = CD$.
Given that $AB = 4 \, cm$,it follows that $CD = 4 \, cm$.

(A) Let the angles of the quadrilateral be $3x, 4x, 4x,$ and $7x$.
The sum of the angles of a quadrilateral is $360^{\circ}$.
Therefore,$3x + 4x + 4x + 7x = 360^{\circ}$.
$18x = 360^{\circ}$.
$x = \frac{360^{\circ}}{18} = 20^{\circ}$.
Now,calculate the individual angles:
First angle $= 3 \times 20^{\circ} = 60^{\circ}$.
Second angle $= 4 \times 20^{\circ} = 80^{\circ}$.
Third angle $= 4 \times 20^{\circ} = 80^{\circ}$.
Fourth angle $= 7 \times 20^{\circ} = 140^{\circ}$.
Thus,the angles are $60^{\circ}, 80^{\circ}, 80^{\circ},$ and $140^{\circ}$.

(N/A) Given: $ABCD$ is a parallelogram,$X$ is the mid-point of $AD$,and $Y$ is the mid-point of $BC$.
$1$. Since $AD \parallel BC$ and $AD = BC$,we have $DX \parallel BY$ and $DX = BY$ (as $X$ and $Y$ are mid-points).
$2$. Therefore,$XBYD$ is a parallelogram because one pair of opposite sides is equal and parallel.
$3$. This implies $PX \parallel QD$ and $PY \parallel BQ$.
$4$. In $\triangle AQD$,$X$ is the mid-point of $AD$ and $XP \parallel QD$. By the Converse of Mid-point Theorem,$P$ is the mid-point of $AQ$. Thus,$AP = PQ$.
$5$. In $\triangle CPB$,$Y$ is the mid-point of $BC$ and $YQ \parallel PB$. By the Converse of Mid-point Theorem,$Q$ is the mid-point of $PC$. Thus,$PQ = QC$.
$6$. From the above two results,we get $AP = PQ = QC$.

(N/A) Given: $ABCD$ is a parallelogram where $AX$ bisects $\angle A$ and $CY$ bisects $\angle C$.
$1$. Since $ABCD$ is a parallelogram,$\angle A = \angle C$ (opposite angles of a parallelogram are equal).
$2$. Dividing both sides by $2$,we get $\frac{1}{2} \angle A = \frac{1}{2} \angle C$.
$3$. Since $AX$ and $CY$ are bisectors,this implies $\angle XAB = \angle YCD$.
$4$. Also,in parallelogram $ABCD$,$AB \parallel DC$. Therefore,$\angle XAB = \angle AXC$ (alternate interior angles).
$5$. From steps $3$ and $4$,$\angle AXC = \angle YCD$.
$6$. Since these are corresponding angles for lines $AX$ and $CY$ with transversal $DC$,$AX \parallel CY$.

(B) The sum of the interior angles of a quadrilateral is $360^{\circ}$.
Let the three equal angles be $x^{\circ}$ each.
Given that one angle is $108^{\circ}$,we can write the equation:
$108^{\circ} + x + x + x = 360^{\circ}$
$108^{\circ} + 3x = 360^{\circ}$
$3x = 360^{\circ} - 108^{\circ}$
$3x = 252^{\circ}$
$x = 252^{\circ} / 3$
$x = 84^{\circ}$
Therefore,each of the three equal angles is $84^{\circ}$.

(N/A) $ABCD$ is a trapezium in which $AB \parallel DC$.
Since $AB \parallel DC$ and $AD$ is a transversal,the sum of interior angles on the same side of the transversal is $180^{\circ}$.
Therefore,$\angle A + \angle D = 180^{\circ}$.
Substituting the value of $\angle A = 45^{\circ}$:
$45^{\circ} + \angle D = 180^{\circ}$
$\angle D = 180^{\circ} - 45^{\circ} = 135^{\circ}$.
Similarly,since $AB \parallel DC$ and $BC$ is a transversal,the sum of interior angles on the same side of the transversal is $180^{\circ}$.
Therefore,$\angle B + \angle C = 180^{\circ}$.
Substituting the value of $\angle B = 45^{\circ}$:
$45^{\circ} + \angle C = 180^{\circ}$
$\angle C = 180^{\circ} - 45^{\circ} = 135^{\circ}$.
Thus,the angles are $\angle C = 135^{\circ}$ and $\angle D = 135^{\circ}$.

(N/A) Let the parallelogram be $ABCD$. Let $DP \perp AB$ and $DQ \perp BC$ be the two altitudes from the obtuse vertex $D$.
In quadrilateral $DPBQ$,the sum of the interior angles is $360^{\circ}$.
$\angle PDQ + \angle DPB + \angle B + \angle DQB = 360^{\circ}$
Given $\angle PDQ = 60^{\circ}$,$\angle DPB = 90^{\circ}$,and $\angle DQB = 90^{\circ}$.
$60^{\circ} + 90^{\circ} + \angle B + 90^{\circ} = 360^{\circ}$
$\angle B + 240^{\circ} = 360^{\circ}$
$\angle B = 360^{\circ} - 240^{\circ} = 120^{\circ}$.
Since opposite angles of a parallelogram are equal,$\angle D = \angle B = 120^{\circ}$.
Since consecutive interior angles are supplementary,$\angle A + \angle B = 180^{\circ}$.
$\angle A + 120^{\circ} = 180^{\circ} \Rightarrow \angle A = 60^{\circ}$.
Since opposite angles are equal,$\angle C = \angle A = 60^{\circ}$.
Thus,the angles of the parallelogram are $60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}$.

(A) Let $P$ be the point on $AB$ such that $DP \perp AB$. Given $AP = PB$.
In $\triangle APD$ and $\triangle BPD$:
$AP = BP$ (Given)
$\angle APD = \angle BPD = 90^{\circ}$ (Altitude)
$PD = PD$ (Common side)
By $SAS$ congruence criterion,$\triangle APD \cong \triangle BPD$.
Therefore,$AD = BD$ $(CPCT)$.
Since $ABCD$ is a rhombus,$AD = AB$. Thus,$AD = BD = AB$.
This implies $\triangle ABD$ is an equilateral triangle.
Therefore,$\angle DAB = 60^{\circ}$.
Since opposite angles of a rhombus are equal,$\angle BCD = 60^{\circ}$.
Since consecutive angles are supplementary,$\angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Similarly,$\angle ADC = 120^{\circ}$.
The angles of the rhombus are $60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}$.

(N/A) Given: $A$ parallelogram $ABCD$. $E$ and $F$ are points on the diagonal $AC$ such that $AE = CF$.
To prove: $BFDE$ is a parallelogram.
Proof: Let the diagonals $AC$ and $BD$ of the parallelogram $ABCD$ intersect at point $O$.
Since the diagonals of a parallelogram bisect each other,we have:
$OA = OC$ ... $(1)$
$OD = OB$ ... $(2)$
Given that $AE = CF$ ... $(3)$
Subtracting equation $(3)$ from equation $(1)$,we get:
$OA - AE = OC - CF$
$OE = OF$ ... $(4)$
Now,in quadrilateral $BFDE$,the diagonals $BD$ and $EF$ bisect each other at point $O$ (from equations $(2)$ and $(4)$).
$A$ quadrilateral whose diagonals bisect each other is a parallelogram.
Therefore,$BFDE$ is a parallelogram.

(N/A) Given: $A$ trapezium $ABCD$ in which $AB \parallel DC$ and $E$ is the mid-point of the side $AD$. Also,$EF \parallel AB$.
To prove: $F$ is the mid-point of $BC$.
Construction: Join $AC$ which intersects $EF$ at $O$.
Proof: In $\triangle ADC$,$E$ is the mid-point of $AD$ and $EF \parallel DC$.
$[\because EF \parallel AB \text{ and } DC \parallel AB \Rightarrow AB \parallel EF \parallel DC]$
$\therefore O$ is the mid-point of $AC$. [Converse of mid-point theorem]
Now,in $\triangle CAB$,$O$ is the mid-point of $AC$ and $OF \parallel AB$.
$\Rightarrow OF$ bisects $BC$ [Converse of mid-point theorem].
Or $F$ is the mid-point of $BC$.
Hence,proved.

(N/A) Given: $\triangle ABC$ and $\triangle PQR$ in which $AB \parallel PQ$,$BC \parallel RQ$ and $CA \parallel PR$.
To prove: $BC = \frac{1}{2} QR$
Proof: Consider the quadrilateral $ARBC$.
Since $AR \parallel BC$ (as $RQ \parallel BC$) and $RB \parallel AC$ (as $PR \parallel AC$),$ARBC$ is a parallelogram.
Therefore,$AR = BC$ (opposite sides of a parallelogram are equal) ... $(1)$
Now,consider the quadrilateral $ABCQ$.
Since $AQ \parallel BC$ (as $RQ \parallel BC$) and $QC \parallel AB$ (as $QP \parallel AB$),$ABCQ$ is a parallelogram.
Therefore,$AQ = BC$ (opposite sides of a parallelogram are equal) ... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$AR + AQ = BC + BC$
$QR = 2BC$
$BC = \frac{1}{2} QR$
Hence,proved.

(N/A) Given: $\triangle ABC$ is an equilateral triangle. $D, E$ and $F$ are the mid-points of the sides $BC, CA$ and $AB$ respectively of $\triangle ABC$.
To prove: $\triangle DEF$ is an equilateral triangle.
Proof: 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.
$1$. $EF$ joins the mid-points of sides $AB$ and $AC$. Therefore,$EF = \frac{1}{2} BC$ $(1)$.
$2$. $DE$ joins the mid-points of sides $BC$ and $AC$. Therefore,$DE = \frac{1}{2} AB$ $(2)$.
$3$. $DF$ joins the mid-points of sides $BC$ and $AB$. Therefore,$DF = \frac{1}{2} AC$ $(3)$.
Since $\triangle ABC$ is an equilateral triangle,we have $AB = BC = CA$ $(4)$.
Substituting $(4)$ into $(1), (2)$ and $(3)$,we get:
$EF = \frac{1}{2} BC$,$DE = \frac{1}{2} BC$,and $DF = \frac{1}{2} BC$.
Thus,$DE = EF = DF$.
Therefore,$\triangle DEF$ is an equilateral triangle. Hence proved.

(N/A) Given: $ABCD$ is a parallelogram,$P$ is on $AB$,$Q$ is on $CD$,and $AP = CQ$.
Let $AC$ and $PQ$ intersect at point $O$.
In $\Delta OAP$ and $\Delta OCQ$:
$1$. $AP = CQ$ (Given)
$2$. $\angle OAP = \angle OCQ$ (Alternate interior angles,as $AB \parallel CD$)
$3$. $\angle AOP = \angle COQ$ (Vertically opposite angles)
Therefore,$\Delta OAP \cong \Delta OCQ$ by $ASA$ congruence rule.
By $CPCT$,$OA = OC$ and $OP = OQ$.
Since $OA = OC$,$O$ is the midpoint of $AC$.
Since $OP = OQ$,$O$ is the midpoint of $PQ$.
Thus,$AC$ and $PQ$ bisect each other.

(N/A) Given: $\angle BAP = \angle DAP = \frac{1}{2} \angle A \quad \dots(1)$
Since $ABCD$ is a parallelogram,we have $\angle A + \angle B = 180^{\circ} \quad \dots(2)$
[Sum of interior angles on the same side of the transversal is $180^{\circ}$]
In $\triangle ABP,$ we have $\angle BAP + \angle B + \angle APB = 180^{\circ}$
Substituting values from $(1)$ and $(2)$:
$\frac{1}{2} \angle A + (180^{\circ} - \angle A) + \angle APB = 180^{\circ}$
$\Rightarrow \angle APB = \frac{1}{2} \angle A \quad \dots(3)$
From $(1)$ and $(3),$ we get $\angle BAP = \angle APB.$
Therefore,$BP = AB$ [Sides opposite to equal angles are equal].
Since opposite sides of a parallelogram are equal,$AD = BC.$
$\Rightarrow \frac{1}{2} AD = \frac{1}{2} BC = BP$ [Since $P$ is the mid-point of $BC$].
$\Rightarrow \frac{1}{2} AD = AB$ [Since $BP = AB$].
Since $AB = CD$ (opposite sides of a parallelogram),we have:
$\frac{1}{2} AD = CD \Rightarrow AD = 2 CD.$
Hence,proved.

(N/A) We draw the figure as per the given conditions.
It is given that $PQ = RS$ and $PQ \parallel RS$. Therefore,$PQSR$ is a parallelogram.
So,$PR = QS$ and $PR \parallel QS$ ... $(1)$
Now,$PR \parallel QS$.
Therefore,$\angle RPQ + \angle PQS = 180^{\circ}$ (Interior angles on the same side of the transversal).
i.e.,$\angle RPQ + \angle PQM + \angle MQS = 180^{\circ}$ ... $(2)$
Also,$PN \parallel QM$ (By construction).
Therefore,$\angle NPQ + \angle PQM = 180^{\circ}$ ... $(3)$
i.e.,$\angle NPR + \angle RPQ + \angle PQM = 180^{\circ}$.
Comparing $(2)$ and $(3)$,we get $\angle NPR = \angle MQS$ ... $(4)$
Similarly,$\angle NRP = \angle MSQ$ ... $(5)$
Therefore,$\Delta PNR \cong \Delta QMS$ [By $ASA$ congruence rule,using $(1)$,$(4)$,and $(5)$].
So,$PN = QM$ and $NR = MS$ $(CPCT)$.
As $PN = QM$ and $PN \parallel QM$,we have $PQMN$ is a parallelogram.
So,$MN = PQ$ and $NM \parallel PQ$.

(N/A) Let $ABCD$ be a parallelogram and $AC$ be a diagonal. Observe that the diagonal $AC$ divides parallelogram $ABCD$ into two triangles,namely,$\Delta ABC$ and $\Delta CDA$. We need to prove that these triangles are congruent.
In $\Delta ABC$ and $\Delta CDA$,note that $BC \parallel AD$ and $AC$ is a transversal.
So,$\angle BCA = \angle DAC$ (Pair of alternate interior angles).
Also,$AB \parallel DC$ and $AC$ is a transversal.
So,$\angle BAC = \angle DCA$ (Pair of alternate interior angles).
And $AC = CA$ (Common side).
Therefore,$\Delta ABC \cong \Delta CDA$ (by $ASA$ congruence rule).
Thus,the diagonal $AC$ divides the parallelogram $ABCD$ into two congruent triangles $ABC$ and $CDA$.

(N/A) Let $ABCD$ be a rhombus and $P, Q, R,$ and $S$ be the mid-points of sides $AB, BC, CD,$ and $DA,$ respectively. Join $AC$ and $BD.$
From triangle $ABD,$ by the mid-point theorem,we have:
$SP = \frac{1}{2} BD$ and $SP \parallel BD.$
Similarly,in triangle $BCD,$ we have:
$RQ = \frac{1}{2} BD$ and $RQ \parallel BD.$
Therefore,$SP = RQ$ and $SP \parallel RQ$ (Equation $1$).
Since one pair of opposite sides is equal and parallel,$PQRS$ is a parallelogram.
Also,the diagonals of a rhombus are perpendicular,so $AC \perp BD.$
In triangle $BAC,$ by the mid-point theorem,$PQ \parallel AC.$
Since $SP \parallel BD$ and $PQ \parallel AC,$ and $AC \perp BD,$ it follows that $SP \perp PQ.$
Thus,$\angle SPQ = 90^{\circ}$ (Equation $2$).
Since $PQRS$ is a parallelogram with one angle equal to $90^{\circ},$ $PQRS$ is a rectangle.

(N/A) Let us consider a parallelogram $ABCD$ where $AC$ is a diagonal that bisects $\angle BAD$.
Given: $\angle BAC = \angle DAC$.
To prove: $\angle BCA = \angle DCA$.
Proof:
Since $AB \parallel CD$ and $AC$ is a transversal,the alternate interior angles are equal:
$\angle BAC = \angle DCA$ --- $(1)$
Since $AD \parallel BC$ and $AC$ is a transversal,the alternate interior angles are equal:
$\angle DAC = \angle BCA$ --- $(2)$
From the given condition,we know:
$\angle BAC = \angle DAC$ --- $(3)$
Comparing equations $(1)$,$(2)$,and $(3)$,we get:
$\angle BCA = \angle DCA$.
Thus,the diagonal $AC$ bisects the opposite angle $\angle BCD$ as well.