Question

If one diagonal of a traperium divides the
diagonal in 1:2 ratio, prove that one of the parallel
sides is double the other.​

Answer 1

Step-by-step explanation:

See the attachment please.....

Answer 2

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HERE IS UR ANSWER

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In the below diagram, we have shown a trapezium ABCD and the two diagonals AC and BD such that both these diagonals intersect at point E.

In the above figure, AB and DC are the parallel sides of the trapezium.

It is given that one of the diagonals of the trapezium divides the other in the ratio of 1:3 so applying this definition in the diagram that we have drawn we get,

$\frac{DE}{EB} = \frac{3}{1}$

We have to prove that DC = 3AB.

To prove that DC = 3AB we are going to first prove the similarity of the triangles DEC and AEB.

$in \: \triangle \: DEC\: and \: \triangle \: AEB$

AB is parallel to DC and AC, BD are transversals so 

$\angle \: DEC \: and \: \angle \:AEB$

are vertically opposite angles and vertically opposite angles are equal to each other.

$\angle \: DEC \: = \: \angle \:AEB$

AB is parallel to DC and BD is transversal so

$\angle \: ABE \: and \: \angle \: EDC$

are alternate interior angles and we know that alternate interior angles are equal to each other.

$\angle \: ABE \: = \: \angle \: EDC$

So by AA similarity condition, we can say that

$\triangle \: DEC \: is \: similar \: to \: \triangle \: AEB$

We know that, if two triangles are similar then ratios of their corresponding sides are equal.

$\frac{DE}{EB} = \frac{DC}{AB} ..(i)$

And we have shown above that:

$\frac{DE}{EB} = \frac{3}{1}$

Substituting the above relation in eq. (1) we get,

$\frac{3}{1} = \frac{DC}{AB}$

On cross-multiplying the above relation, we get,

$3AB = DC$

Hence, we have proved that DC = 3AB