Solution

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

In $\triangle A B C$ and $\triangle D C B$,

$A B=D C$ (Opposite sides of a parallelogram are equal)

$BC=BC($ Common)

$AC=DB$ (Given)

$\therefore \triangle ABC \triangle DCB$ (By SSS Congruence rule)

$\Rightarrow$

$A B C=D C B$

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

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

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

Solution

Let $A B C D$ be a square. Let the diagonals $A C$ and $B D$ 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 $A C=B D, O A=O C, O B=O D$, and $\angle A O B=90^{\circ}$.

In $\triangle A B C$ and $\triangle D C B$,

$A B=D C$ (Sides of a square are equal to each other)

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

$B C=C B$ (Common side)

$\therefore \quad \triangle ABC \cong \triangle DCB$ (By SAS congruency)

$\therefore \quad AC=DB(By CPCT)$

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

In $\triangle AOB$ and $\triangle COD$, $\angle AOB=\angle COD$ (Vertically opposite angles)

$\angle ABO=CDO$ (Alternate interior angles)

$A B=C D$ (Sides of a square are always equal)

$\angle \triangle AOB^{\angle} \triangle COD$ (By AAS congruence rule)

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

Hence, the diagonals of a square bisect each other.

In $\triangle AOB$ and $\triangle COB$,

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

$AO=CO$

$A B=C B$ (Sides of a square are equal)

$BO=BO$ (Common)

$\angle \triangle AOB \triangle COB$ (By SSS congruency)

$\angle AOB=COOB(By CPCT)$

However, $\hat{A O B}+\angle C O B=180^{\circ}$ (Linear pair)

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

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

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

Solution

(i) $A B C D$ is a parallelogram.

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

And, $B A C=D C A$ (Alternate interior angles) … (2) However, it is given that $A C$ bisects $\angle A$.

$\angle \angle DAC=\angle BAC \ldots$

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

$\angle D A C=B C A=B A €=\angle D C A$.

$\angle$ DCA $=B C A$

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

(ii)From equation (4), we obtain

$\angle DAC=\angle DCA$

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

However, $D A=B C$ and $A B=C D$ (Opposite sides of a parallelogram)

$\angle A B=B C=C D=D A$ Hence, $A B C D$

is a rhombus.

Solution

(i) It is given that $A B C D$ is a rectangle.

$\angle \angle A=\angle C$

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

$C D=D A$ (Sides opposite to equal angles are also equal)

However, $D A=B C$ and $A B=C D$ (Opposite sides of a rectangle are equal)

$\angle A B=B C=C D=D A$

$A B C D$ is a rectangle and all of its sides are equal.

Hence, $A B C D$ is a square.

(ii) Let us join BD.

In $\triangle B C D$,

$B C=C D$ (Sides of a square are equal to each other)

$\angle C D B=\mathbb{C} B D$ (Angles opposite to equal sides are equal)

However, $́ CDB=\angle ABD$ (Alternate interior angles for $A B | C D$ )

$\angle CBD=ABD$

$\angle BD$ bisects $B$.

Also, $C^{C} B D=A D B$ (Alternate interior angles for $B C | A D$ )

$\angle \angle CDB=A ABD \angle$

BD bisects b.

Solution

(i) In $\triangle APD$ and $\triangle CQB$,

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

$A D=C B$ (Opposite sides of parallelogram ABCD)

$DP=BQ$ (Given)

$\angle \triangle APD \angle \triangle CQB$ (Using SAS congruence rule) ii)

As we had observed that $\triangle APD \angle \triangle CQB,($

$\angle AP=CQ(CPCT)$

(iii) In $\triangle AQB$ and $\triangle CPD$,

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

$A B=C D$ (Opposite sides of parallelogram ABCD)

$B Q=D P$ (Given)

$\angle \triangle AQB \angle \triangle CPD$ (Using SAS congruence rule) iv)

As we had observed that $\triangle AQB \angle \triangle CPD$, (

$\angle AQ=CP(CPCT)$

(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 $\triangle APB$ and $\triangle CQD$,

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

$A B=C D$ (Opposite sides of parallelogram $A B C D) \quad \angle A B P$

$=$ CDQ (Alternate interior angles for $A B || C D$ )

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

(ii) By using the above result

$\triangle APB \angle \triangle CQD$, we obtain $A P=C Q(B y C P C T)$

Solution

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

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

However, $A D=B C$ (Given)

Therefore, $BC=CE$

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

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

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

${ }^{\angle} A+C B E=180^{\circ}$ (Using the relation $\angle C E B=\angle C B E$ ) $\ldots(1)$

However, $\hat{B}+C B E=180^{\circ}$ (Linear pair angles) $\ldots$ (2)

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

$=\angle B$

(ii) $\quad A B || C D$

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

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

$\angle A+D \leqq C+B$

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

(iii) In $\triangle A B C$ and $\triangle B A D$,

$A B=B A($ Common side)

$B C=A D$ (Given)

$\angle B=A$ (Proved before)

$\angle \triangle ABC \triangle BAD$ (SAS congruence rule) (iv) We had observed that, $\triangle ABC \angle \triangle BAD$

$\angle AC=BD(By CPCT)$