Question

In the figure, ABCD Is a parallelogram and p is any point on BC. Prove that ar (ΔABP) +ar(ΔDPC) =ar(ΔPDA).

Answer 1

Solution -

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

Since BC || AD, so distance between them remains the same

AL = PM

ar(ΔABP) +ar(ΔDPC)

$= \frac{1}{2} \times bp \times al + \frac{1}{2} \times pc \times al$


$= \frac{1}{2} \times al \times (bp + pc)$

$= \frac{1}{2} \times al \times bc = \frac{1}{2} \times pm \times ad$


(Al = PM and BC = AD)

= ar ( ΔPDA).


I hope it's help you

Answer 2

Solution -

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).


I hope it's help you