**Let ABC be a triangle in which AB = AC and let I be its in centre. Suppose BC = AB + AI. Find ∠BAC.**

Extend CA to D such that AD = AI , then CD = CB by the hypothesis.

Hence ∠CDB = ∠CBD = 90° - (C/2) using angle sum property in triangle BCD.

Using angle sum property in triangles ABC and ABI we have ∠AIB = 90° + (C/2).

Thus ∠AIB + ∠ADB = 90° - (C/2) + 90° + (C/2) = 180°.

Hence ADBI is a cyclic quadrilateral. This implies that ∠ADI = ∠ABI = (B/2)

Now,triangle ADI is isosceles,as AD = AI , this gives ∠DAI = 180°-2(∠ADI)=180° - B.

Thus ∠CAI = B which gives A = 2B. As ∠C = ∠B , we get 4B = 180° and hence B = 45°.

Thus we get A = 2B = 90°