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 \[\angle A = 90^o \] .

(Regional Mathematics Olympiad, India, 2016)

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).