Question

In quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at point $E$ such that
$AE:EC=BE:ED$.
Show that: $ABCD$ is a trapezium.

Answer 1

Here, $ABCD$ is a quadrilateral with diagonals $AC$ and $BD$ intersect at point $E$.

$\Rightarrow$ $\frac{AE}{EC}=\frac{BE}{ED}$ [ Given ]

$\Rightarrow$ $\frac{AE}{BE}=\frac{EC}{ED}$ ---- ( 1 )

In $△ABE$ and $△CDE,$

$\Rightarrow$ $\frac{AE}{BE}=\frac{EC}{ED}$ ] From ( 1 ) ]

$\Rightarrow$ $\angle ABE=\angle DEC$ [ Vertically opposites angles ]

$\therefore$ $△ABE\sim △CDE$ [ By $SAS$ similarity theorem ]

$\Rightarrow$ $\angle EDC=\angle EBA$ [ Corresponding angles are equal. ]

$\therefore$ $\angle BDC=\angle ABD$

Bu, this is a pair of alternate angles.

$\therefore$ $AB\parallel DC$

Thus, the quadrilateral $ABCD$ is trapezium.