Question

If the diagonals of a quadrilateral divide each other proportionally, then prove that the quadrilateral is a trapezium.

Answer 1

Let $ABCD$ be a quadrilateral as shown in the above figure:

Diagonal $AC$ and $BD$ divide each other proportionately that is,

$\frac{AO}{OC}=\frac{BO}{OD}$

Therefore, the triangles containing these sides are equiangular that is $△AOB$ and $△COD$ and thus,

$\angle OAB=\angle OCD$

$\angle OBA=\angle ODC$ (Alternate angles)

Therefore, $DC||AB$

Hence, $ABCD$ is a trapezium.