Question

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

Answer 1

Given,

$ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ & $\angle C$

Generally, A rectangle is a quadrilateral whose opposite side are parallel and equal & one angle is ${90}^o$

let us now conssider $△ABC$ and $△ADC$

$\angle BAC=\angle DAC$ [$\therefore$ given $AC$ bisects $\angle A$]

$\angle BCA=\angle DCA$ [$\therefore$ given $AC$ bisects $\angle C$]

$AC=AC$ [$\therefore$ common side]

$\therefore △ABC\cong △ADC$ [By $ASA$ congruency rule].

$AB=AD$ and $CB=CD$ [$\therefore$ by corresponding parts of Congurent].

But in a rectangle opposite side are equal i.e, $AB=DC$ and $BC=AD$

$\therefore AB=BC=CD=DA$

So, here we got the given $ABCD$ is a square.

Now, let us consider $△ABD$ & $△CDB$

$AD=CB$ ($\therefore$ side of a square)

$BD=CD$ ($\therefore$ side of a square)

$BD=BD$ ($\therefore$ Common)

$\therefore △ABD\cong △CBD$ [$\therefore$ by $SSS$ congurence rule].

So therefore, $\angle ABD=\angle CBD---\left(1\right)$

$\angle ADB=\angle CDB---\left(2\right)$

$\therefore \left(1\right)$ and $\left(2\right)$ are by corresponding parts of congurent.

$\therefore$ Diagonal $BD$ bisects $\angle B$ as $\angle D$

Hence Proved.