Question

D and E are mid points of AB and AC of ABC BC is produced to P DE DP and EP are joined Prove that area of PED is equal to area of ADE

Answer 1

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).