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:

 

rmo2016problem1

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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