Question

Question 5
ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.

Answer 1

AB = BC (given)
$\therefore$ $\angle 2=\angle 1$ ...(i)
[angles opposite to equal sides are equal]
and AD = CD [given]
$\Rightarrow \angle 4=\angle 3$ ...(ii)
[angles opposite to equal sides are equal]
On adding Eqs. (i) and (ii), we get
$\angle 2+\angle 4=\angle 1+\angle 3\Rightarrow \angle BCD=\angle BAD...\left(iii\right)In\Delta BADand\Delta BCD,$
AB = BC [given]
$\angle BAD=\angle BCD[fromEq.(iii\left)\right]$
and AD = CD [given]
$\therefore \Delta BAD\cong \Delta BCD\left[bySAScongruencerule\right]$
Hence, $\angle ABD=\angle CBDand\angle ADB=\angle CDB$
$i.e.,$ BD bisects the angles ABC and ADC.