Mix Examples - Quadrilaterals Questions in English

(N/A) Let us join $AC$.
In $\Delta ABC$,since $AB$ is the smallest side of the quadrilateral $ABCD$,we have $AB < BC$ and $AB < AC$.
Specifically,$BC > AB$.
Since the angle opposite to the longer side is larger,we have $\angle BAC > \angle BCA$ ........... $(1)$
In $\Delta ACD$,since $CD$ is the longest side of the quadrilateral $ABCD$,we have $CD > AD$ and $CD > AC$.
Specifically,$CD > AD$.
Since the angle opposite to the longer side is larger,we have $\angle CAD > \angle ACD$ ........... $(2)$
Adding $(1)$ and $(2)$,we get:
$(\angle BAC + \angle CAD) > (\angle BCA + \angle ACD)$
$\Rightarrow \angle A > \angle C$
Similarly,by joining $BD$ and using the same logic for $\Delta ABD$ and $\Delta BCD$,we can show that $\angle B > \angle D$.

(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 $156^{\circ}$.

(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 $\angle ADB = 50^{\circ}$.

(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.

(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) 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.

(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.

(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 $EF$.

(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) 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 $60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}$.

(B) We have $ABCD$ as a parallelogram and $X$ as the mid-point of $AB$.
Now,$\text{ar}(ABCD) = \text{ar}(AXCD) + \text{ar}(\triangle XBC) \dots(1)$
Since the diagonal $AC$ of a parallelogram divides it into two triangles of equal area,we have $\text{ar}(ABCD) = 2 \times \text{ar}(\triangle ABC) \dots(2)$
Since $X$ is the mid-point of $AB$,the median $CX$ of $\triangle ABC$ divides it into two triangles of equal area. Thus,$\text{ar}(\triangle XBC) = \frac{1}{2} \text{ar}(\triangle ABC) \dots(3)$
Substituting $(1)$ and $(3)$ into $(2)$,we get:
$2 \times \text{ar}(\triangle ABC) = 24 + \frac{1}{2} \text{ar}(\triangle ABC)$
$2 \times \text{ar}(\triangle ABC) - \frac{1}{2} \text{ar}(\triangle ABC) = 24$
$\frac{3}{2} \text{ar}(\triangle ABC) = 24$
$\text{ar}(\triangle ABC) = \frac{24 \times 2}{3} = 16 \text{ cm}^2$.
Since $16 \text{ cm}^2 \neq 24 \text{ cm}^2$,the given statement is False.

(A) The area of a trapezium is given by the formula: $\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.
Let the length of one parallel side be $x \, cm$.
Then,the length of the other parallel side is $(x + 4) \, cm$.
Given,$\text{Area} = 475 \, cm^2$ and $\text{height} = 19 \, cm$.
Substituting these values into the formula:
$475 = \frac{1}{2} \times (x + x + 4) \times 19$
$475 = \frac{1}{2} \times (2x + 4) \times 19$
$475 = (x + 2) \times 19$
$x + 2 = \frac{475}{19} = 25$
$x = 25 - 2 = 23 \, cm$.
Therefore,the two parallel sides are $23 \, cm$ and $(23 + 4) \, cm$,which are $23 \, cm$ and $27 \, cm$.

(B) The length of the rectangular plot is $40 \, m$ and the width (breadth) is $15 \, m$.
According to the rules, a space of $3 \, m$ must be left at the front and back, and a space of $2 \, m$ must be left on each of the other two sides.
The new length of the construction area $= 40 - (3 + 3) = 40 - 6 = 34 \, m$.
The new width of the construction area $= 15 - (2 + 2) = 15 - 4 = 11 \, m$.
The largest area where the house can be constructed $= \text{Length} \times \text{Width} = 34 \, m \times 11 \, m = 374 \, m^2$.

(N/A) Given: $A$ quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $O$.
To prove: $AB + BC + CD + DA < 2(BD + AC)$.
Proof: In $\triangle AOB$,by the triangle inequality theorem (the sum of any two sides of a triangle is greater than the third side),we have:
$OA + OB > AB$ ... $(1)$
In $\triangle BOC$,we have:
$OB + OC > BC$ ... $(2)$
In $\triangle COD$,we have:
$OC + OD > CD$ ... $(3)$
In $\triangle DOA$,we have:
$OD + OA > DA$ ... $(4)$
Adding equations $(1), (2), (3),$ and $(4)$,we get:
$(OA + OB) + (OB + OC) + (OC + OD) + (OD + OA) > AB + BC + CD + DA$
$2(OA + OC) + 2(OB + OD) > AB + BC + CD + DA$
Since $OA + OC = AC$ and $OB + OD = BD$,we substitute these into the inequality:
$2(AC) + 2(BD) > AB + BC + CD + DA$
$2(AC + BD) > AB + BC + CD + DA$
Therefore,$AB + BC + CD + DA < 2(AC + BD)$.
Hence,proved.

(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: $\triangle ABC$ 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: $\angle BAP = \angle DAP = \frac{1}{2} \angle A \quad \dots(1)$
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) $ABC$ is an isosceles right triangle with $AB = BC$. $A$ square $BFED$ is inscribed in triangle $ABC$ such that $\angle B$ is common and $\angle B = 90^{\circ}$.
In $\triangle ADE$ and $\triangle EFC$,we have:
$DE = EF$ (Sides of a square are equal) ... $(1)$
$\angle 1 + \angle 2 = 180^{\circ}$ (Linear pair axiom)
Since $\angle 1 = 90^{\circ}$ (angle of a square),$\angle 2 = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Similarly,$\angle 4 = 90^{\circ}$.
Therefore,$\angle 2 = \angle 4 = 90^{\circ}$ ... $(2)$
Since $AB = BC$ (given),$\angle A = \angle C$ ... $(3)$ (Angles opposite to equal sides are equal).
From $(1)$,$(2)$,and $(3)$,by $AAS$ congruence rule,$\triangle ADE \cong \triangle EFC$.
Hence,$AE = EC$ by $CPCT$,which means $E$ bisects the hypotenuse $AC$.

(D) $ABCD$ is a parallelogram,in which $AB = 10 \, cm$ and $AD = 6 \, cm$. The bisector of $\angle A$ meets $DC$ in $E$. $AE$ and $BC$ produced meet at $F$.
Since $AE$ is the bisector of $\angle A$,we have $\angle BAE = \angle EAD \quad \dots(1)$.
Since $AD \parallel BC$ and $AF$ is a transversal,$\angle EAD = \angle EFB$ (Alternate interior angles) $\dots(2)$.
From $(1)$ and $(2)$,$\angle BAE = \angle EFB$.
In $\triangle ABF$,since $\angle BAE = \angle EFB$,the sides opposite to these angles are equal,so $BF = AB$.
Given $AB = 10 \, cm$,therefore $BF = 10 \, cm$.
We know $BF = BC + CF$. Since $BC = AD = 6 \, cm$ (opposite sides of a parallelogram are equal),
$10 \, cm = 6 \, cm + CF$.
$CF = 10 \, cm - 6 \, cm = 4 \, cm$.

(N/A) Given: $A$ quadrilateral $ABCD$ in which $AC \perp BD$ and $P, Q, R$ and $S$ are respectively the mid-points of the sides $AB, BC, CD$ and $DA$ of quadrilateral $ABCD$.
To Prove: $PQRS$ is a rectangle.
Proof: In $\Delta ABC$,$P$ and $Q$ are the mid-points of sides $AB$ and $BC$ respectively.
By the Mid-point theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC \dots(1)$
In $\Delta ADC$,$R$ and $S$ are the mid-points of $CD$ and $AD$ respectively.
By the Mid-point theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC \dots(2)$
From $(1)$ and $(2)$,$PQ \parallel SR$ and $PQ = SR$.
Since one pair of opposite sides is equal and parallel,$PQRS$ is a parallelogram.
Now,let $AC$ and $BD$ intersect at $O$. Let $PQ$ intersect $BD$ at $E$ and $PS$ intersect $AC$ at $G$.
Since $PQ \parallel AC$,$PE \parallel OG$.
Since $PS \parallel BD$,$PG \parallel OE$.
Thus,$PGOE$ is a parallelogram.
Since $AC \perp BD$,$\angle EOG = 90^{\circ}$.
In a parallelogram,opposite angles are equal,so $\angle GPE = \angle EOG = 90^{\circ}$.
Since $PQRS$ is a parallelogram with one angle equal to $90^{\circ}$,$PQRS$ is a rectangle.

(N/A) Given: $A$ parallelogram $ABCD$ in which $P$ and $Q$ are the mid-points of the sides $AB$ and $CD$ respectively. $AQ$ intersects $DP$ at $S$ and $BQ$ intersects $CP$ at $R$.
To prove: $PRQS$ is a parallelogram.
Proof: Since $ABCD$ is a parallelogram,$DC \parallel AB$ and $DC = AB$.
Since $P$ is the mid-point of $AB$ and $Q$ is the mid-point of $CD$,we have $AP = \frac{1}{2} AB$ and $QC = \frac{1}{2} CD$.
Since $AB = CD$,it follows that $AP = QC$.
Also,$AP \parallel QC$ because $AB \parallel DC$.
Since one pair of opposite sides ($AP$ and $QC$) is equal and parallel,$APCQ$ is a parallelogram.
Therefore,$AQ \parallel PC$,which implies $SQ \parallel PR$.
Similarly,we can show that $APQD$ is a parallelogram (since $AP \parallel DQ$ and $AP = DQ$),which implies $AQ \parallel DP$. Also,$PBCQ$ is a parallelogram (since $PB \parallel QC$ and $PB = QC$),which implies $BQ \parallel CP$.
In quadrilateral $PRQS$,we have $SQ \parallel PR$ (from $AQ \parallel PC$) and $SP \parallel QR$ (from $DP \parallel BQ$).
Since both pairs of opposite sides are parallel,$PRQS$ is a parallelogram.

(N/A) Given: $A$ quadrilateral $ABCD$ in which $AB \parallel CD$ and $AD = BC$.
To prove: $\angle A = \angle B$ and $\angle C = \angle D$.
Construction: Draw $DP \perp AB$ and $CQ \perp AB$.
Proof: In $\triangle APD$ and $\triangle BQC$,we have:
$\angle 1 = \angle 2 = 90^{\circ}$ (By construction)
$AD = BC$ (Given)
$DP = CQ$ (Distance between parallel lines is constant)
By $RHS$ criterion of congruence,we have:
$\triangle APD \cong \triangle BQC$ (By $CPCT$)
$\therefore \angle A = \angle B$
Now,since $DC \parallel AB$:
$\angle A + \angle D = 180^{\circ}$ (Sum of consecutive interior angles is $180^{\circ}$)
$\angle B + \angle C = 180^{\circ}$ (Sum of consecutive interior angles is $180^{\circ}$)
Since $\angle A = \angle B$,we have:
$180^{\circ} - \angle A = 180^{\circ} - \angle B$
$\Rightarrow \angle D = \angle C$
Hence,proved.

(N/A) Given: $AB \parallel DE$,$AB = DE$,$AC \parallel DF$ and $AC = DF$.
To prove: $BC \parallel EF$ and $BC = EF$.
Proof:
$1$. In quadrilateral $ACFD$,we are given $AC \parallel DF$ and $AC = DF$. Since one pair of opposite sides is equal and parallel,$ACFD$ is a parallelogram.
Therefore,$AD \parallel CF$ and $AD = CF$ (Opposite sides of a parallelogram are equal and parallel).
$2$. In quadrilateral $ABED$,we are given $AB \parallel DE$ and $AB = DE$. 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).
$3$. From the above results,we have $CF \parallel AD$ and $BE \parallel AD$. This implies $CF \parallel BE$.
Also,$CF = AD$ and $BE = AD$. This implies $CF = BE$.
$4$. In quadrilateral $BCFE$,we have $CF \parallel BE$ and $CF = BE$. Since one pair of opposite sides is equal and parallel,$BCFE$ is a parallelogram.
Therefore,$BC \parallel EF$ and $BC = EF$ (Opposite sides of a parallelogram are equal and parallel).
Hence,proved.

(N/A) Given: $A$ parallelogram $ABCD$ in which the bisectors of angles $\angle A, \angle B, \angle C, \angle D$ intersect at points $P, Q, R, S$ to form a quadrilateral $PQRS$.
To Prove: $PQRS$ is a rectangle.
Proof: Since $ABCD$ is a parallelogram,we have $AB \parallel DC$.
Since $AB \parallel DC$ and the transversal $AD$ intersects them at $A$ and $D$ respectively,the sum of consecutive interior angles is $180^{\circ}$.
Therefore,$\angle A + \angle D = 180^{\circ}$.
Dividing by $2$,we get $\frac{1}{2} \angle A + \frac{1}{2} \angle D = 90^{\circ}$.
Since $AS$ and $DS$ are the bisectors of $\angle A$ and $\angle D$ respectively,we have $\angle DAS + \angle ADS = 90^{\circ} \quad \dots(1)$.
In $\triangle DAS$,by the angle sum property of a triangle,$\angle DAS + \angle ADS + \angle ASD = 180^{\circ}$.
Substituting $(1)$ into this,we get $90^{\circ} + \angle ASD = 180^{\circ}$,which implies $\angle ASD = 90^{\circ}$.
Since $\angle PSR$ and $\angle ASD$ are vertically opposite angles,$\angle PSR = \angle ASD = 90^{\circ}$.
Similarly,we can prove that $\angle SRQ = 90^{\circ}$,$\angle RQP = 90^{\circ}$,and $\angle SPQ = 90^{\circ}$.
Since all angles of the quadrilateral $PQRS$ are $90^{\circ}$,$PQRS$ is a rectangle.

(N/A) $ABCD$ is a parallelogram. Its diagonals $AC$ and $BD$ bisect each other at $O$. $PQ$ passes through the point of intersection $O$ of its diagonals $AC$ and $BD$.
In $\triangle AOP$ and $\triangle COQ$,we have:
$\angle 3 = \angle 4$ [Alternate interior angles,as $AD \parallel BC$]
$OA = OC$ [Diagonals of a parallelogram bisect each other]
$\angle 1 = \angle 2$ [Vertically opposite angles]
Therefore,$\triangle AOP \cong \triangle COQ$ [By $ASA$ congruence rule]
So,$OP = OQ$ [$CPCT$]
Hence,$PQ$ is bisected at $O$.

(N/A) Given: $A$ rectangle $ABCD$ in which diagonal $BD$ bisects $\angle B$.
To prove: $ABCD$ is a square.
Proof: $DC \parallel AB$ [Since opposite sides of a rectangle are parallel].
$\Rightarrow \angle 4 = \angle 1$ ... $(1)$ [Alternate interior angles].
Similarly,$\angle 3 = \angle 2$ ... $(2)$ [Alternate interior angles].
And $\angle 1 = \angle 2$ ... $(3)$ [Given,as $BD$ bisects $\angle B$].
From equations $(1)$,$(2)$,and $(3)$,we get $\angle 3 = \angle 4$.
In $\triangle ABD$ and $\triangle CBD$,we have:
$\angle 1 = \angle 2$ [Given]
$BD = BD$ [Common side]
$\angle 3 = \angle 4$ [Proved above]
So,by $ASA$ criterion of congruence,$\triangle ABD \cong \triangle CBD$.
Therefore,$AB = BC$ [Corresponding parts of congruent triangles].
Since $ABCD$ is a rectangle with adjacent sides equal,$ABCD$ is a square.
Hence,proved.

(N/A) $ABCD$ is a parallelogram. $P$ is the mid-point of $CD$. $A$ line through $C$ parallel to $PA$ intersects $AB$ at $Q$ and $DA$ produced at $R$.
In $\Delta DCR$,$P$ is the mid-point of $CD$ and $AP \parallel CR$ (since $AP \parallel CQ$ and $Q$ lies on $CR$).
Therefore,by the converse of the Mid-point Theorem,$A$ is the mid-point of $DR$,which implies $DA = AR$.
Now,consider $\Delta ARQ$ and $\Delta BCQ$:
$1$. $AR = BC$ (Since $AD = AR$ as proved above,and $AD = BC$ as opposite sides of a parallelogram).
$2$. $\angle 1 = \angle 2$ (Vertically opposite angles).
$3$. $\angle 3 = \angle 4$ (Alternate interior angles,as $AB \parallel DC$ and $RC$ is a transversal).
Therefore,$\Delta ARQ \cong \Delta BCQ$ by the $ASA$ congruence rule.
Thus,$CQ = QR$ by $CPCT$ (Corresponding Parts of Congruent Triangles).
Hence,$DA = AR$ and $CQ = QR$ is proved.

(N/A) Let $ABCD$ be a convex quadrilateral.
Draw a diagonal $AC$ which divides the quadrilateral into two triangles,$\triangle ABC$ and $\triangle ADC$.
We know that the sum of the angles of a triangle is $180^{\circ}$.
For $\triangle ABC$,$\angle BAC + \angle ABC + \angle BCA = 180^{\circ}$ (Equation $1$).
For $\triangle ADC$,$\angle DAC + \angle ADC + \angle DCA = 180^{\circ}$ (Equation $2$).
Adding Equation $1$ and Equation $2$,we get:
$(\angle BAC + \angle DAC) + \angle ABC + \angle ADC + (\angle BCA + \angle DCA) = 180^{\circ} + 180^{\circ}$.
From the figure,$\angle BAC + \angle DAC = \angle A$ and $\angle BCA + \angle DCA = \angle C$.
Therefore,$\angle A + \angle B + \angle C + \angle D = 360^{\circ}$.
Hence,the sum of the angles of a convex quadrilateral is $360^{\circ}$.

(N/A) In a triangle,the sum of any two sides is always greater than the third side.
Consider the triangles formed by the diagonals of the quadrilateral $ABCD$:
$1$. In $\triangle ABC$,$AB + BC > AC$ (Equation $1$).
$2$. In $\triangle ADC$,$AD + CD > AC$ (Equation $2$).
$3$. In $\triangle ABD$,$AB + AD > BD$ (Equation $3$).
$4$. In $\triangle BCD$,$BC + CD > BD$ (Equation $4$).
Adding Equations $1$ and $2$ gives: $(AB + BC + AD + CD) > 2AC$.
Adding Equations $3$ and $4$ gives: $(AB + AD + BC + CD) > 2BD$.
Adding these two resulting inequalities: $2(AB + BC + CD + DA) > 2(AC + BD)$.
Dividing by $2$,we get: $AB + BC + CD + DA > AC + BD$.

(N/A) Let $O$ be the intersection point of the diagonals $AC$ and $BD$.
In $\triangle ABC$,by the triangle inequality,$AB + BC > AC$.
In $\triangle ADC$,by the triangle inequality,$CD + DA > AC$.
Adding these two inequalities,we get: $(AB + BC + CD + DA) > 2AC$.
Similarly,in $\triangle ABD$,$AB + AD > BD$.
In $\triangle BCD$,$BC + CD > BD$.
Adding these two inequalities,we get: $(AB + AD + BC + CD) > 2BD$.
Now,consider the triangles formed by the intersection point $O$:
In $\triangle OAB$,$AB < OA + OB$.
In $\triangle OBC$,$BC < OB + OC$.
In $\triangle OCD$,$CD < OC + OD$.
In $\triangle ODA$,$DA < OD + OA$.
Summing these four inequalities: $AB + BC + CD + DA < 2(OA + OB + OC + OD)$.
Since $OA + OC = AC$ and $OB + OD = BD$,we have $AB + BC + CD + DA < 2(AC + BD)$.

(N/A) Given : $A$ parallelogram $ABCD$ such that the bisectors of consecutive angles $A$ and $B$ intersect at $P$.
To Prove : $\angle APB = 90^{\circ}$.
Proof : Since $ABCD$ is a parallelogram,$AD \parallel BC$.
Now,$AD \parallel BC$ and transversal $AB$ intersects them. Therefore,$\angle A + \angle B = 180^{\circ}$ (Sum of consecutive interior angles is $180^{\circ}$).
Dividing by $2$,we get $\frac{1}{2} \angle A + \frac{1}{2} \angle B = 90^{\circ}$.
Since $AP$ is the bisector of $\angle A$ and $BP$ is the bisector of $\angle B$,we have $\angle PAB + \angle PBA = 90^{\circ}$.
In $\Delta APB$,the sum of angles is $180^{\circ}$:
$\angle PAB + \angle APB + \angle PBA = 180^{\circ}$.
Substituting the value,$90^{\circ} + \angle APB = 180^{\circ}$.
Therefore,$\angle APB = 180^{\circ} - 90^{\circ} = 90^{\circ}$.

(N/A) In quadrilateral $PQRS$,the sum of all interior angles is $360^{\circ}$.
Therefore,$\angle P + \angle Q + \angle R + \angle S = 360^{\circ}$.
In $\triangle PMQ$,the sum of angles is $180^{\circ}$.
Thus,$\angle PMQ + \angle MPQ + \angle MQP = 180^{\circ}$.
Since $PM$ and $QM$ are bisectors,$\angle MPQ = \frac{1}{2} \angle P$ and $\angle MQP = \frac{1}{2} \angle Q$.
Substituting these into the triangle equation: $\angle PMQ + \frac{1}{2} \angle P + \frac{1}{2} \angle Q = 180^{\circ}$.
Multiplying by $2$,we get $2 \angle PMQ + \angle P + \angle Q = 360^{\circ}$.
From the quadrilateral sum,$\angle P + \angle Q = 360^{\circ} - (\angle R + \angle S)$.
Substituting this into the equation: $2 \angle PMQ + 360^{\circ} - (\angle R + \angle S) = 360^{\circ}$.
Simplifying,we get $2 \angle PMQ = \angle R + \angle S$.

(A) Let the angles of the quadrilateral be $2x, 4x, 5x,$ and $7x$ degrees.
We know that the sum of the interior angles of a quadrilateral is $360^{\circ}$.
Therefore,$2x + 4x + 5x + 7x = 360^{\circ}$.
$18x = 360^{\circ}$.
$x = 20^{\circ}$.
Now,calculating each angle:
$\angle A = 2 \times 20^{\circ} = 40^{\circ}$.
$\angle B = 4 \times 20^{\circ} = 80^{\circ}$.
$\angle C = 5 \times 20^{\circ} = 100^{\circ}$.
$\angle D = 7 \times 20^{\circ} = 140^{\circ}$.
Since $\angle B + \angle C = 80^{\circ} + 100^{\circ} = 180^{\circ}$,the adjacent angles on side $BC$ are supplementary,which implies $AB \parallel CD$. $A$ quadrilateral with one pair of parallel sides is a trapezium.

(A) Let the angles of the quadrilateral be $x, 4x, 2x,$ and $2x$ respectively.
We know that the sum of the interior angles of a quadrilateral is $360^{\circ}$.
Therefore,$x + 4x + 2x + 2x = 360^{\circ}$.
$9x = 360^{\circ}$.
$x = 360^{\circ} / 9 = 40^{\circ}$.
Now,calculating each angle:
$\angle A = x = 40^{\circ}$.
$\angle B = 4x = 4 \times 40^{\circ} = 160^{\circ}$.
$\angle C = 2x = 2 \times 40^{\circ} = 80^{\circ}$.
$\angle D = 2x = 2 \times 40^{\circ} = 80^{\circ}$.
Thus,the angles are $40^{\circ}, 160^{\circ}, 80^{\circ},$ and $80^{\circ}$.