Diagonals AC and BD of a trapezium ABCD intersect at O, where

Q) Diagonals AC and BD of a trapezium ABCD intersect at O, where AB || DC. If DO/OB = 1/2, then show that AB = 2CD

Ans: Let’s make a diagram for the given question:

In Δ AOB and Δ COD,

∠ ABO = ∠ CDO (being traversal alternate angles with AB ǁ CD)

similarly ∠ BAO = ∠ DCO

Therefore Δ AOB ~ Δ COD

We know that when 2 triangles are similar, the ratio of corresponding sides is equal

∴ $\frac{AB}{CD} = \frac {OB}{DO}$ ……. (i)

since it is given that DO/OB = 1/2

∴ OB = 2 DO ……….. (ii)

Transferring this value of OB from equation (ii) in equation (i), we get:

∴ $\frac{AB}{CD} = \frac {2 DO}{DO}$ = 2

Therefore, AB = 2 CD

Hence Proved !

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