AB is a diameter of a circle and C is any point on the circumference of the circle. Then,

In △ABC, we havesin⁡θ=BCAB and cos⁡θ=ACAB⇒ BC=ABsin⁡θ and AC=ABcos⁡θLet ∆ be the area of △ABC. Then,Δ=12BC×AC=12(AB)2sin⁡θcos⁡θ=14(AB)2sin⁡2θClearly, it is maximum when sin⁡2θ=1 is maximum i.e. sin⁡2θ=1. In that case, θ=π/4∴ BC=AC=AB2Hence, the triangle is isoceles,