Question

plzzz give me solution​

Answer 1

To Prove :

• $\sf{ \frac{AO}{BO} = \frac{CO}{DO} }$

Given :

• AB || DC

In the given Trapezium,

Let us construct a line EF that is parallel to AB.

Have a look at the attachment!

We know that,

AB || CB,

So, EF || AB.

And, EF || CD.

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In $\sf{ \triangle{ADC}}$,

EO || DC.

According to Basic Proportionality Theorem,

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

_______________

In $\sf{ \triangle{ABD}}$,

EO || AB.

According to Basic Proportionality Theorem,

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

_________________

From (1) and (2),

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

Rearranging the terms,

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

$\sf{ \red{Hence \: Proved!}}$

_________________

Ab kya hee dekh rhe ho?

Chalo itna help krdiya maine, Thanks to de hee skte ho naa? :tears: xD xD

Have a nice day! :D

Answer 2

Given ;

A trapezium "ABCD" in which AB || CD

To prove :

AO =

$\frac{oc}{od}$

BO