Question 2

the question was

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

Solution:                                

We observe that∠AIB = 90◦ +(C/2). Extend CA to D such that AD = AI.
Then CD =CB by the hypothesis. Hence∠CDB =∠CBD = 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.
But ADI is isosceles, since AD = AI. This gives
∠DAI = 180◦−2(∠ADI)= 180◦−B.
Thus∠CAI = B and this gives A = 2B. Since C = B, we obtain 4B = 180◦ and
hence B = 45◦. We thus get A= 2B = 90◦.


I HOPE YOU LIKED ANY DOUBTS ?