# Ex.8.1 Q8 Quadrilaterals Solution - NCERT Maths Class 9

## Question

\(ABCD\) is a rectangle in which diagonal \(AC\) bisects \(\angle A\)as well as \(\angle C\). Show that:

(i) \(ABCD\) is a square

(ii) Diagonal \(BD\) bisects \(\angle B\) as well as \(\angle D\).

## Text Solution

**What is known?**

\(ABCD\) is a rectangle in which diagonal \(AC\) bisects \(∠A\) as well as \(∠C.\)

**What is unknown?**

How we can show that

(i) \(ABCD\) is a square

(ii) Diagonal \(BD\) bisects \(∠B\) as well as \(∠D.\)

**Reasoning:**

We can use angle bisector property and isosceles triangle property to show given rectangle as square and alternate interior angles property to show BD bisects angles B and D.

**Steps:**

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

\[\begin{align}\angle {\rm{A}} &= \angle {\rm{C}}\\\Rightarrow \frac{1}{2}\angle {\rm{A}} &= \frac{1}{2}\angle {\rm{C}}\\\Rightarrow \angle \rm DAC &= \frac{1}{2}\angle {\rm{DCA }} \\({\text{AC bisects }}&\angle {\rm{A \,and }}\angle\,\, {\rm{C}})\end{align}\]

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

However, \(DA = BC\) and \(AB = CD\)

(Opposite sides of a rectangle are equal)

\(AB = BC = CD = DA\)

\(ABCD\) is a rectangle and all the sides are equal.

Hence, \(ABCD\) is a square.

(ii) Let us join \(BD\).

In \(\Delta {BCD}\)

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

\(\angle CDB=\angle CBD \)(Angles opposite to equal sides are equal)

However, \(\angle CDB = \angle ABD\) (Alternate interior angles for \(AB \;||\; CD\))

\[\angle CDB = \angle ABD\]

\(BD\) bisects \(\angle B\)

Also, \(\angle CDB = \angle ABD\) (Alternate interior angles for \(BC \;||\; AD\)) \(\angle CDB = \angle ABD\)

\(BD\) bisects \(\angle D\) and \(\angle B\).