# 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 $\angle$ADC be DE where E is a point on the side AC. $\angle$ADE = $\angle$CDE = $\alpha$ (say). As DE is the tangent at D for the circumcircle of $\triangle$ABD, we get $\angle$DAB = $\angle$ADE = $\alpha$.
Using exterior angle theorem in $\triangle$ABD, we get $\angle$ADC = $\angle$ABD + $\angle$DAB $\Rightarrow$ 2$\alpha$ = $\alpha$ + $\angle$DAB $\Rightarrow$ $\angle$DAB = $\alpha$. This makes $\triangle$ABD isosceles with BD = AD. So AD = DC as D is the mid-point of BC. Therefore, $\triangle$ADC 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$$\,$ + $\angle DAC = \alpha$ + $\, 90^{\circ}\, -\, \alpha = 90^{\circ}$