Question

Prove that $ar\left(\Delta APB\right)=ar\left(\Delta ACQ\right)$

Answer 1

In a $\Delta AXP$ and $\Delta CXB$,
$\angle PAX=XCB$ (alternative angles AP||BC)
AX=CX(given)
$\angle AXP=\angle CXB$ (vertically opposite angles)
$\Delta \simeq \Delta CXB$ (ASA rule)
$\Rightarrow AP=BC$ (By cpct) .....(1)
Similarly QA=BC .....(2)
From (1) and (2) we get
$AP=QA$
Now, AP||BC and AP=QA
$ar\left(\Delta APB\right)=ar\left(\Delta ACQ\right)$ [Triangles having equal bases and between the same parallels QP and BC].