Question

in quadrilateral ABCD diagonal AC and BD intersect each other at E such that AE:EC = BE : ED show that ABCD trapezium

Answer 1

The answer for the above question is given in the picture above.
[Thank You]

Answer 2

Therefore, $ABCD$ is a trapezium.

Step-by-step explanation:

Given;

A quadrilateral $ABCD$ and diagonal $AC$ and $BD$ intersect each other at $E$

Such as $AE:EC=BE:ED$

Let,

Draw a parallel line $GH$ with $DC$ on the point $E$

∵ $GH\parallel DC$

∴ $AG:GD=AE:EC$

⇒$\frac{AG}{GD} =\frac{AE}{EC}$ __1

Also,

$EH:DC$

∴$BH:HC=BE:ED$

⇒$\frac{BH}{HC}=\frac{BE}{ED}$__2

From equation-1 & 2,

$\frac{AG}{GD} =\frac{AE}{EC}=\frac{BH}{HC}=\frac{BE}{ED}$

⇒$\frac{AG}{GD} =\frac{BH}{HC}$ (∵ $AE:EC=BE:ED$ )

∴$AB\parallel GH\parallel DC$

∴$AB\parallel DC$

Thus, $AB\parallel DC$ in $ABCD$ quadrilateral.

∴ $ABCD$ is a trapezium.