Question

In the following figure, ABCD is a trapezium in which AB || DC. The diagonals AC and DB intersect at O.

1. Prove that ΔOAB ∼ ΔОСD
2. If OA = 3x − 1, OB = 2x + 1, OC = x + 3 and OD = 5x + 1, find the value of x.

Answer 1

Since AB || DC, and diagonals AC and DB intersect at O:

We have:

• ∠OAB = ∠OCD because AB || DC AC is a transversal.
• ∠OBA = ∠ODC because AB || DC DB is a transversal.

Therefore, by AA similarity,

△OAB ∼ △OCD

From similarity, `(OA)/(OC) = (OB)/(OD)`

Substitute the given values:

`(3x - 1)/(x + 3) = (2x + 1)/(5x + 1)`

(3x − 1)(5x + 1) = (2x + 1)(x + 3)    ...[Cross-multiplied]

15x² − 2x − 1 = 2x² + 7x + 3

13x² − 9x − 4 = 0

13x² − 13x + 4x − 4 = 0

13x(x − 1) + 4(x − 1) = 0

(x − 1)(13x + 4) = 0

x = 1 or x = `-4/13`

Since lengths must be positive:

If x = `-4/13`, then OA = 3x − 1 < 0, not possible.

Thus, x = 1

Hence, △OAB ∼ △OCD and x = 1.