Brazil
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(◻APQD) = 12ar(||gm ABCD)
Answer:
- We know that diagonal AC of ||gm ABCD divides it into two triangles of equal area. ∴
- 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}
