Question

Question 10
The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{BO}=\frac{CO}{DO}$. Show that ABCD is a trapezium.

Answer 1

Given, quadrilateral ABCD in which diagonals AC and BD intersect each other at O such that $\frac{AO}{BO}=\frac{CO}{DO}$.
To Prove that ABCD is a trapezium.
Construction: Through O, draw line EO, where EO || AB, which meets AD at E.
Proof
In $\Delta DAB$, we have EO || AB
$\therefore \frac{DE}{EA}=\frac{DO}{OB}...\left(i\right)$ [By using Basic Proportionality Theorem]
Also, $\frac{AO}{BO}=\frac{CO}{DO}$ (Given)
$\Rightarrow \frac{DO}{OB}=\frac{CO}{AO}.....\left(ii\right)$
From equation (i) and (ii), we get
$\frac{DE}{EA}=\frac{CO}{AO}$
Therefore, by using converse of Basic Proportionality Theorem, EO || DC also EO || AB $\Rightarrow AB||DC$.
Hence, quadrilateral ABCD is a trapezium with AB || CD.