Question

in fig AP||BQ||CR prove that ar(triangle AQC)=ar(triangle PBR)​

Answer 1

Given,

AP || BQ

|| CR

To Prove,

ar(∆AQC)

= ar(∆PBR)

Proof:

ar(△AQB) = ar(△PBQ) — (i)

(On the

same base BQ and between the same parallels AP and BQ.)

also,

ar(△BQC) = ar(△BQR) — (ii)

 (On the same base BQ and between the same

parallels BQ and CR.)

Adding (i) and (ii),

ar(△AQB) +

ar(△BQC) =

ar(△PBQ) +

ar(△BQR)

 ar(△ AQC) = ar(△ PBR)