Question

If one diagonal of a trapezium divides the other diagonal in the ratio 1:2. Prove that one of the parallel sides is double the other

Answer 1

$\Large \bf \bold \purple {➪Given:}$

One of the diagnol of a trapezium divides the other diagnol in the ratio 1:2.

$\Large \bf \bold \purple {➪To~prove:}$

We have to prove that one of the parallel sides is double the other that is, CD = 2AB.

$\Large \bf \bold \purple {➪Proof:}$

In ∆AOB and ∆COD,

• <AOB = <COD [Vertically opposite angles]
• <ABO = <OCD [Alternate angles]

$\sf \green {\therefore~ \triangle AOB~∼ \triangle DOC}~~ [AA~similarity]$

$\mapsto \sf \dfrac{AO}{OD} = \cfrac{AB}{CD}$

[If two triangles are similar then the ratios of their corresponding sides are equal]

$\mapsto \sf \dfrac{1}{2} = \cfrac{AB}{CD}$

$\mapsto \sf CD =2AB$

⭐Hence, proved !!

_____________________

All done :)

Answer 2

Given:

ABCD is a trapezium. ABllCD. Diagonal BC divides the diagonal AC in the ratio 1:2 at O. That is OA:OD = 1:2

To prove:

CD = 2AB

Proof:

In ∆ AOB and ∆ COD,

<AOB = <COD (vertically opposite angles)

<ABO = <OCD ( Alternate angles )

$∴  \: △AOB∼△DOC  \: \: \: \: \: \:  [AA similarity]$

$➠ \frac{ AO }{ OD } = \frac{AB}{CD}$

$➠ \: \: \: \: \frac{1}{2} = \frac{AB}{CD} \: \: \: \:$

$➠ CD = 2AB$

Hence,Proved.