If one diagonal of a trapezium divides the other diagonal in the ratio 1 : 3. Prove that one of the parallel sides is three times the other.

Given: BD divides the diagonal AC in ratio 1:3

To Prove: DC = 3 AB

Proof:

In, Δ AOB and Δ COD

∠ OAB = ∠ OCD [ Alternate angles are equal as AB || CD]

∠ABO = ∠ ODC [Alternate angles are equal as AB || CD]

Therefore, Δ AOB ≈ Δ COD By AA Similarity

Now as both the triangles are similar we can say that,

As

[ Given]

We can say that

And therefore by cross multiplying, CD = 3 AB

Hence, Proved.