Question

$\Delta$ABC and $\Delta$DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that $\Delta$ABD $\cong \Delta$ ACD.

Answer 1

$\Delta ABC$ and $\Delta ACD$
$AB=AC,\Delta ABC$ is isosceles
$BD=DC,\Delta BDC$ is isosceles
$AD=AD$ common
By SSS congruency rule
$\Delta ABC\cong \Delta ACD$