Question

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

A
CP,AP,BP
B
BP,AP,CP
C
AP,BP,CP
D
BP,CP,AP

Solution

## The correct option is B $$BP, AP, CP$$In $$\Delta ACP,$$ we have$$\angle ACP > \angle CAP$$$$\Rightarrow AP > CP...(i)$$In $$\Delta ABP,$$ we have$$\angle BAP > \angle ABP \Rightarrow BP > AP ...(ii)$$From $$(i)$$ and $$(ii),$$ we have$$BP > AP > CP.$$Mathematics

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