In figure, AP || BQ || CR. Prove that ar(ΔAQC) = ar(ΔPBR).

Given: AP || BQ || CR.
To Prove: ar(ΔAQC) = ar(ΔPBR).
Proof: #8757; ΔABAQ and ΔBPQ are on the same base BQ and between the same parallels BQ and AP. ∴ ar(ΔBAQ) = ar(ΔBPQ)    ...(1)
V    Two triangles on the same base (or equal bases) and between the
same parallels are equal in area ∵ ΔBCQ and ΔBQR are on the same base BQ and between the same parallels BQ and CR.
∴ ar(ΔBCQ) = ar(ΔBQR)    ...(2)
V    Two triangles on the same base (or equal bases) and between the
same parallels are equal in area Adding the corresponding sides of (1) and (2), we get
ar(ΔBAQ) + ar(ΔBCQ)
= ar(ΔBPQ) + ar(ΔBQR)
⇒ ar(ΔAQC) = ar(ΔPBR).