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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 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

Grade:11

1 Answers

Shamik Banerjee
27 Points
5 years ago
Let the bisector of \angleADC be DE where E is a point on the side AC. \angleADE = \angleCDE = \alpha (say). As DE is the tangent at D for the circumcircle of \triangleABD, we get \angleDAB = \angleADE = \alpha.
 
Using exterior angle theorem in \triangleABD, we get \angleADC = \angleABD + \angleDAB \Rightarrow 2\alpha = \alpha + \angleDAB \Rightarrow \angleDAB = \alpha. This makes \triangleABD isosceles with BD = AD. So AD = DC as D is the mid-point of BC. Therefore, \triangleADC is isosceles with AD = AC \Rightarrow \angle DAC = \angle DCA = \frac{180^{\circ} - 2\alpha }{2} = 90^{\circ} - \alpha
 
\angle A = \angle BAC = \angle DAB\, + \angle DAC = \alpha + \, 90^{\circ}\, -\, \alpha = 90^{\circ}

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