A geometry problem by Kshetrapal Dashottar
In triangle ABC, let J be the centre of the excircle tangent to side BC at and to the extensions of sides AC and AB at and , respectively. Suppose that the lines and AB are perpendicular and intersect at D. Let E be the foot of the perpendicular from to line DJ. Determine the angles in degrees.
The answer is 90.
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1 solution
Consider the circles ω1, ω2 and ω3 of diameters C1D, A1B and AB1, respectively. Line segments JC1, JB1 and JA1 are tangents to those circles and, due to the right angle at D, ω2 and ω3 pass through point D. Since ∠C1ED is a right angle, point E lies on circle ω1, therefore JC2 1 = JD · JE. Since JA1 = JB1 = JC1 are all radii of the excircle, we also have JA2 1 = JD · JE and JB2 1 = JD · JE. These equalities show that E lies on circles ω2 and ω3 as well, so ∠BEA1 = ∠AEB1 = 90◦