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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



From the above diagram

given that

according to the converse of BPT.

from ΔABC and ΔDEP


then EP parallel to AB

we can write that 

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

and from the same triangles 


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.


SO, ar(ΔDPE) = ar(ΔADE).
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