Problem: Let be a triangle with circumcircle and incenter Let the internal angle bisectors of meet in respectively. Let intersect at and in Let intersect in Suppose the quadrilateral is a kite; that is, and Prove that is an equilateral triangle.
Since PIRQ is a kite, by side-side-side congruence rule. Hence
Also (same angle).
Hence are equiangular. This implies
But (because both of them are subtended by the same segment BC’).
Since ABC is isosceles (BA = BC) and BR is the angle bisector, hence it is perpendicular to AC. Thus .
Join AB’. since they are subtended by the same segment AC’ and CC’ is the angle bisector of C.
Similarly since they are subtended by the same segment BC’ and CC’ is the angle bisector of C.
Since earlier we found ,
therefore is isosceles with AB’=IB’.
Finally AI = IB’ (because since earlier we proved them to be equiangular, and we can also show that AR = B’P; How? Clearly as QP= QR, two right angles equal, vertically opposite angles equal. Thus AQ = QB’. QR = QP’. Adding these two we get AR = PB’).
Since AI = IB’ and earlier we showed IB’ = AB’ hence the triangle is equilateral, implying . Since ABC is an isosceles triangle with one angle hence it is equilateral.