Question

In quadrilateral ABCD, diagonals AC and BD intersect at point E. Such that AE : EC = BE : ED. Then, ABCD is a parallelogram.

• A. True
• B. False

Answer 1

The correct option is A

True

Given: In quadrilateral ABCD, diagonals AC and BD intersect each other at E and AE : EC = BE : ED

$\frac{AE}{EC}=\frac{BE}{ED}\Rightarrow \frac{AE}{BE}=\frac{EC}{ED}$

In $\Delta AEB$ and $\Delta CED$

$\frac{AE}{BE}=\frac{EC}{ED}$ (given)

$\angle AEB=\angle CED$ (Vertically opposite angles)

$\therefore \Delta AEB\sim \Delta CED$ (SAS axiom)

$\therefore \angle EAB=\angle ECD$

$\angle EBA=\angle EDC$

But, these are pairs of alternate angles

$\therefore AB\parallel CD\dots \dots \left(i\right)$

Similarly we can prove that

$AD\parallel BC\dots \dots \left(ii\right)$

$\therefore$ from (i) and (ii)

ABCD is a parallelogram.