Question

Question 9
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $\frac{AO}{BO}=\frac{CO}{DO}$.

Answer 1

Given, ABCD is a trapezium in which AB || DC in which diagonals AC and BD intersect each other at O.
To Prove that $\frac{AO}{BO}=\frac{CO}{DO}$
Construction: Through O, draw EO || DC || AB
Proof
In $\Delta ADC$, we have OE || DC (By Construction)
$\therefore \frac{AE}{ED}=\frac{AO}{CO}...\left(i\right)$ [By using Basic Proportionality Theorem]
In $\Delta ABD$, we have OE || AB (By Construction)
$\therefore \frac{DE}{EA}=\frac{DO}{BO}...\left(ii\right)$ [By using Basic Proportionality Theorem]
From equation (i) and (ii), we get
$\frac{AO}{CO}=\frac{BO}{DO}$
$\therefore \frac{AO}{BO}=\frac{CO}{DO}$