Prove the required condition — task. Mathematics CBSE, Class 10.

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Answer variants:

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

$\angle AOB \sim \triangle COD$

$\frac{BO}{DO}$

[Since corresponding sides of similar triangles]

[Since they are alternate angles]

$AB$ is parallel to

$CD$

$\angle DCO$

$\frac{CO}{DO}$

$\angle OBA$

$ABCD$ is a trapezium. Here, the sides

$AB$ and

$CD$ are parallel to each other. Also, the diagonals intersect at

$O$ . Prove that

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

Let us look at the figure given below for a better understanding.

We already know that in

$\triangle ABC$ and

$\triangle COD$ ,

. [Given]

This makes,

$\angle OAB$

$=$

and

$=$

$\angle ODC$ .

By similarity criterion, we have, "If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar."

Thus,

Also,

$\frac{AO}{CO}$

$=$

which can also be written as

$\frac{AO}{BO}$

$=$

Hence, the required condition

is proved.

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16. Prove the required condition

Exercise condition: