Question

In $\Delta$ABC, $\angle$B$={45}^o$, $\angle$C$={65}^o$ and the bisector of $\angle$BAC meets BC at P. Then the ascending order of sides is ___________.

• A. AP, BP, CP
• B. AP, CP, BP
• C. BP, AP, CP
• D. None of these

Answer 1

The correct option is D None of these

$\angle BAP=\angle CAP={35}^o$ (because AP is angle bisector)

$\angle B>\angle BAP\Rightarrow AP>BP$

and $\angle C>\angle CAP\Rightarrow AP>CP$

$\angle ABC<\angle ACP\Rightarrow CP<BP$

therefore ascending order is $CP,BP,AP$