Question

The diagonals of a quadrilateral $ABCD$ intersects each other at the point $O$ such that $\frac{AO}{BO}=\frac{CO}{DO}$. Show that $ABCD$ is a trapezium.

Answer 1

Construction : $Mo\parallel AB$

Given : $\frac{AO}{BO}=\frac{CO}{DO}$

$\frac{AO}{CO}=\frac{BO}{DO}$ ----- (i)

In $\Delta ABD$

$MO\parallel AB$

SO, $\frac{DM}{AM}=\frac{DO}{BO}$ (by CPCT)

$\frac{AM}{DM}=\frac{AO}{CO}$ ---- (ii) from (i)

Now, in triangle $ADC$

$\frac{AM}{DM}=\frac{AO}{CO}$ from (ii)

So, by converse of B.P.T.

$MO\parallel DC$

Since, $AB\parallel MO$

So, $AB\parallel DC$

Hence, $ABCD$ is trapezium.