In the given figure, D is the midpoint of side AB of ΔABC and P is any point on BC. If CQ||PD meets AB in Q, prove that ar(ΔBPQ) is equal to 12ar(ΔABC).
Answer:
- We are given that D is the midpoint of side AB of ΔABC and P is any point on BC.
Also, CQ||PD meets AB in Q. - Let us join CD and PQ.
- We know that a median of a triangle divides it into two triangles of equal area.
In ΔABC, CD is a median. ∴ ar(ΔBCD)=12ar(ΔABC)⟹ar(ΔBPD)+ar(ΔDPC)=12ar(ΔABC)…(i) - But, ΔDPC and ΔDPQ being on the same base DP and between the same parallels DP and CQ, we have: ar(ΔDPC)=ar(ΔDPQ)…(ii) Using (i) and (ii), we get: ar(ΔBPD)+ar(ΔDPQ)=12ar(ΔABC)∴ ar(ΔBPQ)=12ar(ΔABC)