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.
  4. But, \Delta DPC and \Delta DPQ being on the same base DP and between the same parallels DP and CQ, we have: \begin{aligned} ar(\Delta DPC) = ar(\Delta DPQ) &&\ldots \text{(ii)} \end{aligned} Using \text{(i)} and \text{(ii)}, we get: \begin{aligned} &ar(\Delta BPD) + ar(\Delta DPQ) = \dfrac { 1 } { 2 } ar(\Delta ABC) \\ \therefore \space & ar(\Delta BPQ) = \dfrac { 1 } { 2 } ar(\Delta ABC) \end{aligned}

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