In trapezium ABCD, side AB is parallel to side DC. Diagonals AC and BD intersect at point P. Prove that triangles APD and BPC are equal in area.

Given: ABCD be the trapezium with side AB is parallel to side DC. P is the intersection point of diagonals AC and BD.

To prove: Triangles APD and BPC are equal in area.

Proof: If two triangles lie on the same base and are between the same parallels, their areas are equal.

In the trapezium ABCD, the triangles Δ ACD and Δ BCD share the same base CD and are between the same parallels AB ∥ DC.

Thus, their areas are equal:

Ar.(Δ ACD) = Ar.(Δ BCD)

The diagonals AC and BD intersect at P, dividing the trapezium into smaller triangles. Subtract the area of △DPC, which is common to both △ACD and △BCD, from both sides:

⇒ Ar.(Δ ACD) - Ar.(Δ DPC) = Ar.(Δ BCD) - Ar.(Δ DPC)

⇒ Ar.(Δ APD) = Ar.(Δ BCP)

Hence, the triangles Δ APD and Δ BPC are equal in area.