Question

Question 1 (i)
$\Delta$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 (see the given figure). If AD is extended to intersect BC at P, show that
$\Delta ABD\cong \Delta ACD$

Answer 1

In $\Delta ABD$ and $\Delta ACD$,
AB = AC (given)
BD = CD (given)
AD = AD (common)
$\therefore \Delta ABD\cong \Delta ACD$ (by SSS congruence rule)