Triangle ABC has incentre 'I' . Points X , Y are located on the line segments AB , AC
respectively so that BX * AB = IB ^2 , CY *AC = IC^2 . IF :-
X , I , Y are collinear , find measure of angle 'A'
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Consider the circumcircle of triangle AXI. Since BX *AB=IB², IB is tangent to the aforementioned circum-circle and thus by the alternate segment theorem /BXI=/BAI=/A/2. Likewise /CIY=/A/2. Also if AD is the internal bisector then /BID=A/2+B/2 and /CID=A/2+C/2. Hence we can say that 4 * A/2+2(A/2+B/2+A/2+C/2)=360° Or 4 * A/2+2(A/2+A/2+90-A/2)=360° which yields A=60