Question

Question 1 (ii)
$\Delta$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
$\Delta ABP\cong \Delta ACP$

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)
$\Rightarrow \angle BAD=\angle CAD\dots \left(i\right)$

In $\Delta ABP$ and $\Delta ACP$
AB = AC (given).
$\angle BAP$ = $\angle CAP$ [From (i)]
AP = AP (common)
$\therefore \Delta ABP\cong \Delta ACP$ (by SAS congruence rule)