Two Trinangle having the same base and equal areas lie between the same parallels.

Explanation

Given: Two `triangleABC` and `trianglePBC` on the same base `BC` and between the same parallel lines `BC` and `AP`.
To prove: `ar(∆ABC)=ar(∆PBC)`
Construction: Through `B`, draw `BD` parallel to `CA` intersecting `PA` produced in `D` and through `C`, draw `CQ` parallel to `BP`, intersecting line `AP` in `Q`.
Proof:
We have, `BD` parallel `CA` [By construction]
And, `BC` parallel `DA` [Given]
`∴`Quadrilateral `BCAD` is a parallelogram.
Similarly, Quadrilateral `BCQP` is a parallelogram.
Now, parallelogram `BCQP` and `BCAD` are on the same base `BC`, and between the same parallels.
`∴ ar(BCQP)=ar(BCAD)` ....(i)
We know that the diagonals of a parallelogram divides it into two triangles of equal area.
`∴ar(∆PBC)=1/2ar(BCQP)` ....(ii)
And,
`ar(∆ABC)=1/2ar(BCAD)` ....(iii)
Now,
`ar(BCQP)=ar(BCAD)` [From (i)]
`⇒1/2ar(BCAD)=1/2ar(BCQP)`
Hence, `ar(∆ABC)=ar(∆PBC)` [Using (ii) and (iii)]
Hence Proved.