India
In the given figure, ^@ABCD^@ is a parallelogram whose diagonals ^@AC^@ and ^@BD^@ intersect at ^@O^@. A line segment through ^@O^@ meets ^@AB^@ at ^@P^@ and ^@DC^@ at ^@Q^@.
Prove that
^@ar(\square APQD)^@ = ^@\dfrac { 1 } { 2 } ar(||gm\space ABCD)^@
Answer:
- We know that diagonal ^@AC^@ of ^@||gm \space ABCD^@ divides it into two triangles of equal area. @^ \begin{aligned} \therefore ar(\Delta ACD) = \dfrac { 1 } { 2 } ar(||gm \space ABCD) &&\ldots \text{(i)} \end{aligned} @^
- Now, In ^@\Delta OAP^@ and ^@\Delta OCQ^@, we have: @^ \begin{aligned} &OA = OC &&[\text{Diagonals of a ||gm bisect each other}] \\ &\angle AOP = \angle COQ &&[\text{Vertically opposite angles}]\\ &\angle PAO = \angle QCO &&[\text{Alternate interior angles}]\\ \therefore &\Delta OAP \cong \Delta OCQ \end{aligned} @^
- We know that if two triangles are congurrent then their respective areas are equal. @^ \begin{aligned} \therefore& \space ar(\Delta OAP) = ar(\Delta OCQ) \\ \implies& ar(\Delta OAP) + ar( \square AOQD) = ar(\Delta OCQ) + ar(\square AOQD) \\ \implies& ar(\square APQD) = ar(\Delta ACD) \\ &\space\space\space\space\space\space\space\space\space\space\space\space\space \space\space\space\space\space\space\space\space\space = \dfrac { 1 } { 2 } ar(||gm\space ABCD) &&[\text{Using eq (i)}] \\ \therefore&\space ar(\square APQD) = \dfrac { 1 } { 2 } ar(||gm\space ABCD) \end{aligned} @^
