Question

Prove that the diagonals of a trapezium divide each other proportionally.

Answer 1

Given :

A trapezium ABCD in which $AB\parallel DC$ and its diagonals AC and BD intersect at O.

To prove :

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

Construction :

Through O, draw $EO\parallel AB$, meeting AD at E.

Proof :

In $△ADC,EO\parallel DC$

Therefore,

$\frac{AE}{ED}=\frac{AO}{OC}$ .........(1)

In $△DAB,EO\parallel AB$

Therefore,

$\frac{AE}{ED}=\frac{BO}{OD}$ ..........(2)

From 1 and 2, we get,

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

Hence, proved.