Question

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

1. $ABCD$ is a square.
2. Diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

Answer 1

Step $1:$ Drawing the diagram:

$ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as$\angle C$.

Join diagonal $BD$.

Step $2:$ Proving $ABCD$ is a square:

As, diagonal $AC$ bisects $\angle A$ as well as$\angle C$,

$\begin{array}{rcl}\therefore \angle BAC& =& \angle DAC\\ \angle BCA& =& \angle DCA................\left(1\right)\end{array}$

As we know that every rectangle is a parallelogram.

$ABCD$ is a parallelogram.

$\angle BCA=\angle DAC...................\left(2\right)$ [ Alternate interior angles are equal]

From $\left(1\right)$ and $\left(2\right)$, we have

$\angle DCA=\angle DAC...................\left(3\right)$

In $∆ADC$,

$\angle DCA=\angle DAC$ then,

$AD=CD$ [Sides opposite to equal angles of a triangle are equal]

Similarly, $AB=BC$

In, rectangle $ABCD$

$AB=CD,AD=BC$ (opposite sides of rectangle are equal)

$\therefore AB=BC=CD=DA$

Therefore, $ABCD$ is a square.

Step $3:$ Proving Diagonal $BD$ bisects $\angle B$ as well as $\angle D$:

In $∆ADB$,

$AD=AB$ (Side of square)

$\therefore \angle ABD=\angle ADB$ (Angles opposite to equal sides are equal)

In $∆CDB$,

$CD=CB$ (Side of square)

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

And, $\angle ABD=\angle CDB$ [ Alternate interior angles are equal]

$\begin{array}{rcl}\therefore \angle ABD& =& \angle CBD\\ \angle ADB& =& \angle CDB\end{array}$

Therefore, Diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

Hence proved.