Question

Question 1 (iii)
$\Delta$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
AP bisects $\angle$A as well as $\angle$D.

Answer 1

AB = AC (given)
BD = CD (given)
AD = AD (common)
$\therefore \Delta ABD\cong \Delta ACD$ (by SSS congruence rule)
$\Rightarrow \angle BAD=\angle CAD$ (by CPCT)
$\Rightarrow \angle BAP=\angle CAP$
Hence, AP bisects $\angle A$.

$\angle BDA=\angle CDA\dots \left(i\right)$ [CPCT]
Subtracting (i) from ${180}^{\circ }$,
${180}^{\circ }-\angle BDA={180}^{\circ }-\angle CDA$
$\Rightarrow \angle BDP=\angle CDP$ [$\because$ Linear pair]
$\therefore$ AP bisects $\angle D$ as well.