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Analytical Geometry> let ABC be a triangle and D be the midpoi...

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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

Sundar balarka , 8 Years ago

Grade 11

1 Answers

Shamik Banerjee

Last Activity: 7 Years ago

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 $AD = AC \Rightarrow \angle DAC = \angle DCA = \frac{180^{\circ} - 2\alpha }{2} = 90^{\circ} - \alpha$

$\angle A = \angle BAC = \angle DAB$ $\,$ + $\angle DAC = \alpha$ + $\, 90^{\circ}\, -\, \alpha = 90^{\circ}$

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