In the adjoining figure, ABCD is a parallelogram. Points P and Q on BC trisect BC. Prove that
`ar(triangleAPQ)=ar(triangleDPQ)=(1)/(6)ar(triangleABCD)`.

Explanation

Through P and Q, we draw PM and QN parallel to AB. Parallelograms ABPM, MPQN and NQCD have equal areas since they have equal bases and lie between the same parallels [ Theorem 3].

`therefore ar("||gm ABPM")=ar("||gm MPQN")`
`=ar("||gm NQCD") = (1)/(3)ar("||gm ABCD")....(i)`
Now, `ar(triangleAPQ)=ar(triangleDPQ)`
=`(1)/(2)ar("||gm MPQN")`
[`therefore triangleAPQ, triantgleDPQ` and "||gm" MPQN have the same base PQ and lie between the same parallels AD and BC]
`rArr ar(triangleAPQ)=ar(triangleDPQ)=(1)/(2){(1)/(3)ar("||gm ABCD")}` [using (i)].
Hence, `ar(triangleAPQ)=ar(triangleDPQ)=(1)/(6)ar("||gm ABCD")`.