Question

In the figure X and Y are the mid points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that $ar\left(△ABP\right)=ar\left(△ACQ\right).$

Answer 1

Given: In $△ABC,$ X and Y are the mid points of AC and AB respectively. Through A, a line parallel to BC is drawn. Join BX and CY and produce them to meet the parallel line through A, at P and Q respectively and intersect each other at O.

To prove : ar $\left(△ABP\right)=ar\left(△ACQ\right)$
Construction : Join XY and produce it to both sides
Proof:$\because$ X and Y are mid points of sides AC and AB
$\therefore XY||BC$

Similarly, XY || PQ
$△BXYand△CXY$ are on the same base XY and between the same parallels
$\therefore ar\left(△BXY\right)=ar\left(△CXY\right)$
Now trap. XYAP and trap. XYAQ are on the same base XY and between the same parallels
$\therefore ar\left(XYAP\right)=ar\left(XYAQ\right)$
Adding (i) and (ii),
$ar\left(△BXY\right)+ar\left(XYAP\right)=ar\left(CXY\right)+ar\left(XYAQ\right)\Rightarrow ar\left(△ABP\right)=ar\left(△ACQ\right)$