Question

In the given figure, $\angle BDC=\angle ADC$ and $\angle BCA=\angle ADB$. Show that : (i) $△ACD\cong △BDC$ (ii) $BC=AD$ (iii) $\angle A=\angle B$.

Answer 1

Let AD and BC intersects at P

Now, in $△APC$ & $△BPD$

$\angle ACP=\angle BDP$ (given)

$\angle BPD=\angle APC$ (vertically opposite angles)

So, $\angle B=\angle A$ (angle of triangle)

In $△ACD$ & $△BDC$

$\angle C=\angle D$ (Given)

$\angle B=\angle A$ (proved above)

$CD=CD$ (common)

So, $△ACD\cong △BDC$ (AAS criteria)

So, $BC=AD$ (CPCT)