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in Triangles by (24.0k points)

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

(a) 45º 

(b) 90º 

(c) 60º 

(d) 75º

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1 Answer

+1 vote
by (23.5k points)
edited by

(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.

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