Solution

Let $ABCD$ be a parallelogram. To show that $ABCD$ is a rectangle, we have to prove that one of its interior angles is ${90}^{\circ }$.

In $\mathrm{△}ABC$ and $\mathrm{△}DCB$,

$AB=DC$ (Opposite sides of a parallelogram are equal)

$BC=BC($ Common)

$AC=DB$ (Given)

$\therefore \mathrm{△}ABC\mathrm{△}DCB$ (By SSS Congruence rule)

$\Rightarrow$

$ABC=DCB$

It is known that the sum of the measures of angles on the same side of transversal is ${180}^{\circ }$.

$\begin{array}{cc}\mathrm{\angle }\mathrm{\angle }& ABC+DCB={180}^{\circ }(AB||CD)\\ \Rightarrow \mathrm{\angle }& ABC+\mathrm{\angle }ABC={180}^{\circ }\\ \Rightarrow 2\mathrm{\angle }& ABC={180}^{\circ }\\ \Rightarrow \mathrm{\angle }& ABC={90}^{\circ }\end{array}$

Since $ABCD$ is a parallelogram and one of its interior angles is ${90}^{\circ },ABCD$ is a rectangle.

Solution

Let $ABCD$ be a square. Let the diagonals $AC$ and $BD$ intersect each other at a point $O$. To prove that the diagonals of a square are equal and bisect each other at right angles, we have to prove $AC=BD,OA=OC,OB=OD$, and $\mathrm{\angle }AOB={90}^{\circ }$.

In $\mathrm{△}ABC$ and $\mathrm{△}DCB$,

$AB=DC$ (Sides of a square are equal to each other)

$\mathrm{\angle }ABC=\mathrm{\angle }DCB(.$ All interior angles are of ${90}^{\circ }$ )

$BC=CB$ (Common side)

$\therefore \mathrm{△}ABC\cong \mathrm{△}DCB$ (By SAS congruency)

$\therefore AC=DB(ByCPCT)$

Hence, the diagonals of a square are equal in length.

In $\mathrm{△}AOB$ and $\mathrm{△}COD$, $\mathrm{\angle }AOB=\mathrm{\angle }COD$ (Vertically opposite angles)

$\mathrm{\angle }ABO=CDO$ (Alternate interior angles)

$AB=CD$ (Sides of a square are always equal)

$\mathrm{\angle }\mathrm{△}AO{B}^{\mathrm{\angle }}\mathrm{△}COD$ (By AAS congruence rule)

$\mathrm{\angle }AO=CO$ and $OB=OD$ (By CPCT)

Hence, the diagonals of a square bisect each other.

In $\mathrm{△}AOB$ and $\mathrm{△}COB$,

As we had proved that diagonals bisect each other, therefore,

$AO=CO$

$AB=CB$ (Sides of a square are equal)

$BO=BO$ (Common)

$\mathrm{\angle }\mathrm{△}AOB\mathrm{△}COB$ (By SSS congruency)

$\mathrm{\angle }AOB=COOB(ByCPCT)$

However, $\widehat{AOB}+\mathrm{\angle }COB={180}^{\circ }$ (Linear pair)

$\mathrm{\angle }AOB={180}^{\circ }2$

$\mathrm{\angle }AOB={90}^{\circ }$

Hence, the diagonals of a square bisect each other at right angles.

Solution

(i) $ABCD$ is a parallelogram.

$\mathrm{\angle }DAC=BCA$ (Alternate interior angles) ..

And, $BAC=DCA$ (Alternate interior angles) … (2) However, it is given that $AC$ bisects $\mathrm{\angle }A$.

$\mathrm{\angle }\mathrm{\angle }DAC=\mathrm{\angle }BAC\dots$

From equations (1), (2), and (3), we obtain

$\text{Math input error}$.

$\mathrm{\angle }$ DCA $=BCA$

Hence, $AC$ bisects $C^{\circ }$.

(ii)From equation (4), we obtain

$\mathrm{\angle }DAC=\mathrm{\angle }DCA$

$\mathrm{\angle }DA=DC$ (Side opposite to equal angles are equal)

However, $DA=BC$ and $AB=CD$ (Opposite sides of a parallelogram)

$\mathrm{\angle }AB=BC=CD=DA$ Hence, $ABCD$

is a rhombus.

Solution

(i) It is given that $ABCD$ is a rectangle.

$\mathrm{\angle }\mathrm{\angle }A=\mathrm{\angle }C$

$\begin{array}{rl}& \Rightarrow \frac{1}{2}\mathrm{\angle }A=\frac{1}{2}\mathrm{\angle }C\\ & \Rightarrow \mathrm{\angle }DAC=\mathrm{\angle }DCA(AC\text{ bisects }\mathrm{\angle }A\text{ and }\mathrm{\angle }C)\end{array}$

$CD=DA$ (Sides opposite to equal angles are also equal)

However, $DA=BC$ and $AB=CD$ (Opposite sides of a rectangle are equal)

$\mathrm{\angle }AB=BC=CD=DA$

$ABCD$ is a rectangle and all of its sides are equal.

Hence, $ABCD$ is a square.

(ii) Let us join BD.

In $\mathrm{△}BCD$,

$BC=CD$ (Sides of a square are equal to each other)

$\mathrm{\angle }CDB=\mathbb{C}BD$ (Angles opposite to equal sides are equal)

However, $́CDB=\mathrm{\angle }ABD$ (Alternate interior angles for $AB|CD$ )

$\mathrm{\angle }CBD=ABD$

$\mathrm{\angle }BD$ bisects $B$.

Also, $C^{C}BD=ADB$ (Alternate interior angles for $BC|AD$ )

$\mathrm{\angle }\mathrm{\angle }CDB=AABD\mathrm{\angle }$

BD bisects b.

Solution

(i) In $\mathrm{△}APD$ and $\mathrm{△}CQB$,

$\mathrm{\angle }ADP=\mathrm{\angle }CBQ$ (Alternate interior angles for $BC||AD$ )

$AD=CB$ (Opposite sides of parallelogram ABCD)

$DP=BQ$ (Given)

$\mathrm{\angle }\mathrm{△}APD\mathrm{\angle }\mathrm{△}CQB$ (Using SAS congruence rule) ii)

As we had observed that $\mathrm{△}APD\mathrm{\angle }\mathrm{△}CQB,($

$\mathrm{\angle }AP=CQ(CPCT)$

(iii) In $\mathrm{△}AQB$ and $\mathrm{△}CPD$,

$\mathrm{\angle }ABQ=\mathrm{\angle }CDP$ (Alternate interior angles for $AB||CD$ )

$AB=CD$ (Opposite sides of parallelogram ABCD)

$BQ=DP$ (Given)

$\mathrm{\angle }\mathrm{△}AQB\mathrm{\angle }\mathrm{△}CPD$ (Using SAS congruence rule) iv)

As we had observed that $\mathrm{△}AQB\mathrm{\angle }\mathrm{△}CPD$, (

$\mathrm{\angle }AQ=CP(CPCT)$

(v) From the result obtained in (ii) and (iv),

$AQ=CP$ and $AP=CQ$

Since opposite sides in quadrilateral APCQ are equal to each other, APCQ is a parallelogram.

Solution

(i) In $\mathrm{△}APB$ and $\mathrm{△}CQD$,

$\mathrm{\angle }APB=\mathrm{\angle }CQD(.$ Each ${.90}^{\circ })$

$AB=CD$ (Opposite sides of parallelogram $ABCD)\mathrm{\angle }ABP$

$=$ CDQ (Alternate interior angles for $AB||CD$ )

${}^{\mathrm{\angle }}\mathrm{△}APB\mathrm{\angle }\mathrm{△}CQD$ (By AAS congruency)

(ii) By using the above result

$\mathrm{△}APB\mathrm{\angle }\mathrm{△}CQD$, we obtain $AP=CQ(ByCPCT)$

Solution

Let us extend $AB$. Then, draw a line through $C$, which is parallel to $AD$, intersecting $AE$ at point $E$. It is clear that AECD is a parallelogram.

(i) $AD=CE$ (Opposite sides of parallelogram AECD)

However, $AD=BC$ (Given)

Therefore, $BC=CE$

$\mathrm{\angle }CEB=/CBE$ (Angle opposite to equal sides are also equal)

Consider parallel lines AD and CE. AE is the transversal line for them.

$\mathrm{\angle }A+CEB={180}^{\circ }$ (Angles on the same side of transversal)

${}^{\mathrm{\angle }}A+CBE={180}^{\circ }$ (Using the relation $\mathrm{\angle }CEB=\mathrm{\angle }CBE$ ) $\dots (1)$

However, $\hat{B}+CBE={180}^{\circ }$ (Linear pair angles) $\dots$ (2)

From equations (1) and (2), we obtain $\mathrm{\angle }A$

$=\mathrm{\angle }B$

(ii) $AB||CD$

$\mathrm{\angle }A+D={180}^{\circ }$ (Angles on the same side of the transversal)

Also, $C^c+B={180}^{\circ }$ (Angles on the same side of the transversal)

$\mathrm{\angle }A+D\leqq C+B$

However, $\stackrel{`}{A}=B$ [Using the result obtained in (i) $]\mathrm{\angle }C=$ D

(iii) In $\mathrm{△}ABC$ and $\mathrm{△}BAD$,

$AB=BA($ Common side)

$BC=AD$ (Given)

$\mathrm{\angle }B=A$ (Proved before)

$\mathrm{\angle }\mathrm{△}ABC\mathrm{△}BAD$ (SAS congruence rule) (iv) We had observed that, $\mathrm{△}ABC\mathrm{\angle }\mathrm{△}BAD$

$\mathrm{\angle }AC=BD(ByCPCT)$