Q12 of 25 Page 9

In ΔABC, D is the mid-point of AB. P is any point of BC. CQ||PD meets AB in Q. Show that ar (Δ BPQ) = 1/2 ar(Δ ABC)

Construction: Join CD.


Proof:


Since D is the mid-point of AB. So, in Δ ABC, CD is the median.


So,


ar (Δ BCD) = 1/2 ar(Δ ABC) ….. (1)


Since triangles PDQ and PDC are on the same base PD and between same parallel lines PD and QC.


∴ ar(Δ PDQ) = ar(Δ PDC)


From (1),


ar (Δ BCD) = 1/2 ar(Δ ABC)


⇒ ar (Δ BPD) + ar(Δ PDC)= 1/2 ar(Δ ABC)


⇒ ar (Δ BPD) + ar(Δ PDQ) = 1/2 ar(Δ ABC)


⇒ ar (Δ BPQ) = 1/2 ar(Δ ABC)


Hence Proved


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