Question
In quadrilateral $A B C D$ (See figure). $A C=A D$ and $A B$ bisects $\angle A$. Show that $\triangle A B C \cong \triangle A B D$. What can you say about $B C$ and $B D$ ?
✓
Answer
Given: In quadrilateral $ABCD, AC = AD$ and $AB$ bisects $\angle A.$
To prove: $\angle ABC \cong \triangle ABD$
Proof: In $\triangle A B C$ and $\triangle A B D$,
$A C=A D$ [Given]
$\angle B A C=\angle B A D[\because AB$ bisects $\angle A]$
$AB = AB$ [Common]
$\therefore \triangle ABC \cong \triangle ABD$ [By $SAS$ congruency]
Thus $BC = BD$ [By $C.P.C.T.]$
To prove: $\angle ABC \cong \triangle ABD$
Proof: In $\triangle A B C$ and $\triangle A B D$,
$A C=A D$ [Given]
$\angle B A C=\angle B A D[\because AB$ bisects $\angle A]$
$AB = AB$ [Common]
$\therefore \triangle ABC \cong \triangle ABD$ [By $SAS$ congruency]
Thus $BC = BD$ [By $C.P.C.T.]$
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