In triangle ABC, D is mid-point of AB and P is any point on BC. If CQ parallel to PD meets AB at Q, prove that:

2 x area (△ BPQ) = area (△ ABC)

Given: ABC is a triangle. D is the mid point of AB (BD = DA). P is any point on BC. CQ ∥ PD meets AB at Q.

To Prove: 2 x Ar.ea (△ BPQ) = Ar.ea (△ ABC)

Construction: Join PQ and CD.

Proof: In Δ ABC, D is the mid point of AB.

The line CD is median and a median divides a triangle into two triangles of equal area.

⇒ Ar.(Δ ACD) = Ar.(Δ BCD) = $\dfrac{1}{2}$ Ar.(Δ ABC)

Ar.(Δ BCD) = Ar.(Δ BPD) + Ar.(Δ DPC)

⇒ Ar.(Δ BPD) + Ar.(Δ DPC) = $\dfrac{1}{2}$ Ar.(Δ ABC) …………….(1)

Since CQ ∥ PD and Q lies on AB, the triangles Δ DPQ and Δ DPC have same base (DP) and are between the same parallels (CQ ∥ PD).

∴ Ar.(Δ DPQ) = Ar.(Δ DPC) …………….(2)

So, equation (1) becomes,

⇒ Ar.(Δ BPD) + Ar.(Δ DPQ) = $\dfrac{1}{2}$ Ar.(Δ ABC)

⇒ Ar.(Δ BPQ) = $\dfrac{1}{2}$ Ar.(Δ ABC)

Hence, 2 x area (△ BPQ) = area (△ ABC).