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)
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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