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Question

In ΔABC, B=35, C=65 and the bisector of BAC meets BC in P. Arrange AP, BP and CP in descending order.

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Solution

In ΔABC, B=35, C=65 and AP is the bisector of BAC which meets BC in P.

Arrange PA, PB and PC in descending order

In ΔABC,

A+B+C=180

(Sum of angles of a triangles)

A+35+65=180 A+100=180 A=180100=80 PA is a bisector of BAC 1=2=802=40Now in ΔACP, ACP>CAP C>2 AP>CP ...(i)Similarly, in ΔABP.BAP>ABP 1>B BP>AP ...(ii)From (i) and (ii)BP>AP>CP


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