UPSC CDS 2 2022 Math Paper

Concept:In a triangle with an incircle, the angle between two points of tangency at the third tangency point equals $90^{\circ} - \frac{A}{2}$, where $A$ is the vertex opposite to that angle.Explanation:Let the incircle touch sides BC, CA, AB at D, E, F respectively.Tangents from a common external point are equal: $AF = AE$, $BF = BD$, $CD = CE$.Thus, triangle $AFE$ is isosceles with $AF = AE$.In $\triangle AFE$, $\angle A + \angle AFE + \angle AEF = 180^{\circ}$.Since $\angle AFE = \angle AEF$, we have $A + 2\angle AFE = 180^{\circ}$, so $\angle AFE = 90^{\circ} - \frac{A}{2}$.Now, line AB is tangent at F and chord FE is part of the incircle.By the alternate segment theorem, the angle between tangent AF and chord FE equals the angle in the alternate segment, which is $\angle EDF$ (the angle subtended by chord FE at point D on the circle).Hence, $\angle EDF = \angle AFE = 90^{\circ} - \frac{A}{2}$.Alternatively, using properties of the contact triangle, the angle at D of the contact triangle is directly $90^{\circ} - \frac{A}{2}$.Answer:$90^{\circ} - \frac{A}{2}$, i.e., option D.