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Problem:Let ABC be a triangle in which AB = AC and let Ibe its in-centre. Suppose BC = AB + AI. Find ∠BAC.Solution: Let AB = AC = b, BC = 2aAI = 2a - bAD^2 = AB^2 - BD^2 = b^2 - a^2BI bisects angle ABD in triangle ABD.So AI/ID = AB/BD = b/a and 2a - b = AI = [b/(a + b)].ADHenceb^2 - a^2 = AD^2 = [(2a - b)^2.(a + b)^2]/b^2b^2.(b - a) = (2a - b)^2.(a + b)b^3 - ab^2 = 4a^3 + 4a^2b + ab^2 + b^3 - 4a^2b - 4ab^2= 4a^3 - 3ab^2 + b^3Thus 2ab^2 = 4a^3 , b^2 = 2a^2sin∠BAD = a/b = 1/sqrt 2, ∠BAD = 45 deg∠BAC = 2∠BAD = 90 deg
DEAR SIR I also like math like you .i think that you mase a mistake in above mentioned Q.1 that you write angle DAC in place of angle DAB in equation (ii) in the solution
Thank you Gouravji
Ref.:Soln.of (KVPY Practice) Q.1- equation(ii)Draw CE ∥ DA (meeting AB at E)ADCE is a parallelogramSo CE = DA = CB, ∆CEB is isosceles So ∠CBA = ∠CBE = ∠CEBAlso ∠CEB = ∠DAB (corresp angles, CE ∥ DA)Hence ∠DAB = ∠CBA as stated in (ii)