Question

If one of the diagonal of a trapezium divides the other in the ratio 2:1 prove that one of the parallel sides is twice the other

Answer 1

Given: ABCD is a trapezium. AB||CD. Diagonal AC divides the diagonal BD
in the ratio 2:1 at O. i.e. BO: OD = 2:1

To prove: AB = 2 CD

Proof: In ∆AOB and ∆COD

∠AOB = ∠COD (Vertically opposite angles)
∠ABD = ∠ODC (Alternate angles)

∴ ∆AOB ~ ∆COD (AA similarity criterion)
⇒ AB/CD = AO/OC = BO/OD [Corresponding sides of similar triangles are proportional.]

⇒ AB/CD = BO/OD

⇒ AB/CD =2/1

⇒ AB=2 CD
Hence, if one diagonal of the trapezium divides the other diagonal in the ratio of 2:1,
then one of the parallel sides is double the other.