(b) 90º.
AI and BI are the bisectors of ∠CAB and ∠CBA respectively,
∴ In ΔABC, ∠A + ∠B + ∠C = 180º
⇒ ∠A + ∠B = 180º – ∠C
\(\frac{∠A}{2}+\frac{∠B}{2}=90°-\frac{∠C}{2}\)
Also, in ΔAIB,
∠AIB = 180º – \(\bigg(\frac{∠A}{2}+\frac{∠B}{2}\bigg)\)
= 180º – \(\bigg(90°-\frac{∠C}{2}\bigg)\) = \(90°+\frac{∠C}{2}\)
Extend CA to D such that AD = AI.
Then, BC = AB + AI ⇒ BC = CA + AD = CD
(By hypothesis, AB = AC, AD = AI)
⇒ ∠CDB = ∠CBD = \(90°-\frac{∠C}{2}\)
(Since in ΔCDB, CD = CB and ∠C + ∠D + ∠B = 180°)
Thus, ∠AIB + ∠ADB = \(90°+\frac{∠C}{2}\) + \(90°-\frac{∠C}{2}\) = 180º
⇒ AI BD is a cyclic quadrilateral
Also ∠ADI = ∠ABI = \(\frac{∠B}{2}\) (angles in the same segment)
∴ In ΔDAI, ∠DAI, = 180º – 2 (∠ADI) = 180º – ∠B (∵ AD = AI ⇒ ∠ADI = ∠AID)
Thus, ∠CAI = B ⇒ A = 2B (CAD is a straight angle)
Since, AC = AB ⇒ ∠B = ∠C
∴ In ΔABC, ∠A + ∠B + ∠C = 180º
⇒ 4 ∠B = 180º ⇒ ∠B = 45º ⇒ ∠A = 90º.