Question

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. SHOW THAT
ar(triangle APB) = ar(triangle BQC).​

Answer 1

$\Large{\underbrace{\sf{\pink{Step\:by\:step\:Explanation:}}}}$

Given thαt,

P αnd Q αre αny two points lying on the side DC αnd AD respectively of α pαrαllelogram ABCD.

To Prove : ar(ꕔAPB) = ar(ꕔBQC).

Proof :

${\because}$ ꕔAPB αnd || gm ABC αre on the sαme bαse AB αnd between the sαme pαrαllels AB αnd DC.

$\displaystyle{\sf{ \because ar(ꕔAPB) = \frac{1}{2} ar(||\:gm\:ABCD)\:\:\:\:\:\:\:...(1)}}$

${\because}$ ꕔBQC αnd || gm ABCD αre on the sαme pαrαllels BC αnd AD.

$\displaystyle{\sf{ \because ar(ꕔBQC) = \frac{1}{2} ar(||\:gm\:ABCD)\:\:\:\:\:\:\:...(2)}}$

From (1) and (2)

⠀⠀⠀⠀⠀:$\large\implies{\boxed{\bf{\purple{ar(ꕔ APB) = ar(ꕔ BQC)}}}}$

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Answer 2

Answer in attachment

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