ABCD is a rectangle and P is the mid-point of AB. DP is produced to meet CB at Q. Prove that the area of rectangle ABCD = area of ∆DQC.

In ∆APD and ∆PQB,

AP = BP [Since P is the mid-point of AB]

∠DAP = ∠QBP [each angle is 90° as ABCD is a rectangle]

∠APD = ∠BPQ [Vertically opposite angles are equal]

So, ∆APD ≅ ∆PQB [By using ASA axiom]

So, area of ∆APD = area of ∆PQB …….(i) (As both triangles are congruent.)

From figure,

area of rectangle ABCD = area of ∆APD + area of quad. PBCD

= area of ∆PQB + area of quad. PBCD ……(From i)

= area of ∆DQC.

Hence, proved that area of rectangle ABCD = area of ∆DQC..