Question

In $\Delta ABC,\angle B={35}^{\circ },\angle C={65}^{\circ }\text{and the bisector of}\angle BAC$ meets BC in P. Arrange AP, BP and CP in descending order.

Answer 1

In $\Delta ABC,\angle B={35}^{\circ },\angle C={65}^{\circ }\text{and AP is the bisector of}\angle BAC\text{which meets BC in P.}$

Arrange PA, PB and PC in descending order

In $\Delta ABC,$

$\angle A+\angle B+\angle C={180}^{\circ }$

(Sum of angles of a triangles)

$\Rightarrow \angle A+{35}^{\circ }+{65}^{\circ }={180}^{\circ }\Rightarrow \angle A+{100}^{\circ }={180}^{\circ }\therefore \angle A={180}^{\circ }-{100}^{\circ }={80}^{\circ }\because \text{PA is a bisector of}\angle BAC\therefore \angle 1=\angle 2=\frac{{80}^{\circ }}{2}={40}^{\circ }Nowin\Delta ACP,\angle ACP>\angle CAP\Rightarrow \angle C>\angle 2\therefore AP>CP...\left(i\right)Similarly,in\Delta ABP.\angle BAP>\angle ABP\Rightarrow \angle 1>\angle B\therefore BP>AP...\left(ii\right)\text{From (i) and (ii)}BP>AP>CP$