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