# Orthogonality

Let ABC be a triangle and D be the midpoint of BC. Suppose the angle bisector of $\angle ADC$ is tangent to the circumcircle of triangle ABD at D. Prove that $\angle A = 90^o$ .

Discussion: Since DE is tangent to the circle at D, by tangent-chord theorem, we conclude:

$\angle EDA =\angle ABD$ (angle made by tangent with a chord is equal to the angle subtended by the chord in the alternate segment)…. (i)

But $\angle EDA = \angle EDC$ since DE is the angle bisector of $\angle ADC$. … (ii)

Hence combining (i) and (ii) we have $\angle ABD = \angle EDC$ implying DE is parallel to AB (as corresponding angles are equal).

Then $\angle EDA = \angle DAB$ (alternate angles) … (iii)

Combining (i) and (iii) we conclude $\angle DBA = \angle DAB$.

This implies $\Delta ADB$ is isosceles. Hence DA = DB.

But DB = DC (as D is the midpoint of BC.

Hence DA = DB = DC implying D is a point equidistant from the vertices A, B, C. Hence D is the center of the circumcircle of triangle ABC and BC is the diameter.

There $\angle CAB = 90^o$ (angle in the semicircle).

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