In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that ar(∆ABP) = ar(quad. ABCD).

Question

Philoid · Accepted Answer

Given: AC ||DP

We know that any two or Triangles having the same base and lying between the same parallel lines are equal in area.

∴ Area (Δ ACD) = Area (Δ ACP) –1

Add Area (Δ ABC) on both sides of eq –1

We get,

Area (Δ ACD) + Area (Δ ABC) = Area (Δ ACP) + Area (Δ ABC)

That is,

Area (quad.ABCD) = Area (Δ ABP)