Given: In quadrilateral ABCD, AC = AD and AB bisects $\angle $A. To prove: $\angle $ABC $ \cong$ $\triangle$ABD Proof: In $\triangle$ABC and $\triangle$ABD, AC = AD [Given] $\angle $BAC = $\angle $BAD [$\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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