In the given figure, ∠ACB = 90° and CD ⊥ AB. Prove that

In ΔACB and ΔCDB,

∠ABC = ∠CBD (Common)

∠ACB = ∠CDB (90°)

So, by AA similarity criterion ΔACB ~ ΔCDB

Similarly, In ΔACB and ΔADC,

∠ABC = ∠ADC (Common)

∠ACB = ∠ADC (90°)

So, by AA similarity criterion ΔACB ~ ΔADC

We know that if two triangles are similar then the ratio of their corresponding sides is equal.

⇒ BC² = AB×BD ….(i)

And AC² = AB×AD …..(ii)

Dividing (i) and (ii), we get

Hence, proved.