Question

In the adjacent figure ABCD is a parallelogram and BC is produced to point Q such that AD=CQ. If AQ intersects DC at p. Show that : 1. Ar(ADP) + ar(BPC) =half of parallelogram (ABCD).

Answer 1

Answer:

Step-by-step explanation:

Construction:

Draw DG on BA produced and BF⊥ DC.

Proof:

We have,

Area ΔABD = 1/2 × AB × DG …. (1)

Area ΔBCE = 1/2 × CE × BF ….. (2)

Since Area ΔABD and Area ΔBCE are between same parallels so their heights are equal

i.e. DG = BF

Also,

AB = CE (given)

1/2 × AB × DG = 1/2 × CE × BF

∴ from (1) and (2),

Area ΔABD = Area ΔBCE