Two similar triangles ABC and PBC are made on opposite sides of the

same base BC. Prove that AB = BP.

Since ΔABC∼ΔPBC

⇒∠ABC = ∠PBC,∠ACB =∠PCB and ∠CAB = ∠BPC  [ similar triangles are equiangular ]

Side BC is common in both triangles.

In Δ ABC and ΔPBC

∠ABC = ∠PBC   [GIVEN]

   BC =BC            [ COMMON ]

∠ACB =∠PCB     [GIVEN]

 Δ ABC is congruent to ΔPBC   [ASA]

⇒ AB = BP [ cpct ]