In the given figure, ^@D^@ is the midpoint of side ^@AB^@ of ^@\Delta ABC^@ and ^@P^@ is any point on ^@BC^@. If ^@CQ||PD^@ meets ^@AB^@ in ^@Q^@, prove that ^@ar(\Delta BPQ)^@ is equal to ^@\dfrac { 1 } { 2 } ar(\Delta ABC)^@.
A C B Q D P


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Answer:


Step by Step Explanation:
  1. We are given that ^@D^@ is the midpoint of side ^@AB^@ of ^@\Delta ABC^@ 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 ^@\Delta ABC^@, ^@CD^@ is a median. @^ \begin{aligned} \therefore \space & ar(\Delta BCD) = \dfrac { 1 } { 2 } ar(\Delta ABC) \\ \implies& ar(\Delta BPD)+ar(\Delta DPC) = \dfrac { 1 } { 2 } ar(\Delta ABC) &&\ldots \text{(i)} \end{aligned} @^
  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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