Question

In the fig. ABCD is a quadrilateral. A line through D parallel to AC meets BC produced at E. prove that $ar\left(\Delta ABE\right)=arquad.\left(ABCD\right)$

Answer 1

$ar\left(\Delta ACD\right)=ar\left(\Delta ACE\right)$

both triangles are on same base AC and between same parallel AC and DE

Adding $ar\left(\Delta ABC\right)$ on both sides

$ar\left(\Delta ABC\right)+ar\left(\Delta ACD\right)=ar\left(\Delta ABC\right)+ar\left(\Delta ACE\right)$

$ar\left(quadABCD\right)=ar\left(\Delta ABE\right)$

$\therefore$ Hence proved.