let ABC be a triangle and D be the midpoint of BC. suppose the angle bisector of angle ADC is tangent to the circumcirle of triangle ABD at D.prove that angle A is equal to 90°.RMO 2016

Let the bisector of ADC be DE where E is a point on the side AC. ADE = CDE = (say). As DE is the tangent at D for the circumcircle of ABD, we get DAB = ADE = .

 

Using exterior angle theorem in ABD, we get ADC = ABD + DAB 2 = + DAB DAB = . This makes ABD isosceles with BD = AD. So AD = DC as D is the mid-point of BC. Therefore, ADC is isosceles with

 

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