ABCD is a quadrilateral in which …

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Sia ? 6 years, 2 months ago
AD = BC [Given]
{tex} \angle {/tex}DAB = {tex} \angle {/tex}CBA [Given]
AB = AB [Common]
{tex}\therefore \triangle ABD \cong \triangle BAC{/tex} [By SAS congruency]
ΔABD≅ΔBAC
{tex}\therefore {/tex} BD=AC[By C.P.C.T.]
ΔABD≅ΔBAC{tex}\triangle \mathrm{ABD} \cong \triangle \mathrm{BAC}{/tex}
{tex}\therefore{/tex}{tex} \angle {/tex}ABD = {tex} \angle {/tex}BAC [By C.P.C.T.]
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