Question

In the given figure, ABCD is a parallelogram and Pis
a point on BC. Prove that : ar(ABP) + ar(DPC)
= ar(APD).
​

Answer 1

Answer:

HEY MATE HERE IS THE ANS

Step-by-step explanation:

Draw AL_|_ BC and PM _|_ AD.

BC || AD,

AL = PM

ar(ΔABP) +ar(ΔDPC)

= 1/2*BP *Al+1/2*PC*AL= 1/2*AL*(BP+PC)

= 1/2*AL*BC= 1/2*PM*AD

(Al = PM and BC = AD)

= ar ( ΔPDA).

HOPE THIS HELPS YOU

THANKS❤️✌

Answer 2

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draw a line || to CD as EP