Question

ABCD is a trapezium in which AB||BC and its diagonals intersect each other at a point O. show that
AO/BO=CO/DO​

Answer 1

Given :-

• ABCD is a trapezium in which AB || BC and its diagonals intersect each other at a point O

To prove :-

$\to\rm \frac{AO}{BO} = \frac{CO}{DO}$

Construction :-

• Let us draw a line EF || AB || DC passing through point O.

Proof :-

Now, in △ADC

EO || DC

So,

$\rm\frac{AE}{DE} = \frac{AO}{CO} - - - \: equation \: (1)$

Similarly,

In △DBA

EO || AB ( Because EF || AB )

$\rm\frac{AE}{DE} = \frac{BO}{DO} \: - - - \: equation \: (2)$

From equation(1) and equation(2), we get

$\rm\frac{AO}{CO} = \frac{BO}{DO}$

$\rm\frac{AO}{BO} = \frac{BO}{CO}$

HENCE PROVED

Answer 2

☯️ GIVEN :-

• AB || BC
• Its diagonals intersect each other at a point O.

☯️ TO PROVE :-

AO/BO = CO/DO.

☯️ PROOF :-

We will draw a line such that from O, It will become EF || AB || DC.

So,

In ADC,

EO || DC.

So, We can say,

AE/BE = AO/CO.

Similiar Case with DBC,

So, We can say,

AE/DE = BO/DO.

From these highlighted Equations, We Can Say,

AO/BO = CO/DO.

Hence, Proved.