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).
A C B Q D P


Answer:


Step by Step Explanation:
  1. 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.
  2. Let us join CD and PQ.
    A C B Q D P
  3. 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)
  4. 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)

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