Answers

2016-02-21T19:30:14+05:30
From the above diagram

given that

according to the converse of BPT.

from ΔABC and ΔDEP

CE/EF = CP/PB

then EP parallel to AB

we can write that 

EP parallel to AD ---------------(1)

and from the same triangles 

BD/EA = BP/PC

then DP parallel to AC

we can write that 

DP parallel to AE --------------(2)

from(1) and(2)

ADEP is aparallelogram

so, parallelogram divides the two triangles in two similar triangles.

then ΔDPE≈ΔADE

SO, ar(ΔDPE) = ar(ΔADE).
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