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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
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B
BP,AP,CP
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C
AP,BP,CP
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D
BP,CP,AP
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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
$$ BP > AP > CP.$$

1397024_1669933_ans_889f1ab0cdf74cb78be7b40a5314a321.png

Mathematics

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