Problem:  Let ABC be a triangle with circumcircle \Gamma and incenter I. Let the internal angle bisectors of \angle A,\angle B,\angle C meet \Gamma in A',B',C' respectively. Let B'C' intersect AA' at P, and AC in Q. Let BB' intersect AC in R. Suppose the quadrilateral PIRQ is a kite; that is, IP=IR and QP=QR. Prove that ABC is an equilateral triangle.

Discussion: 

Screen Shot 2015-12-06 at 10.41.28 PM

Since PIRQ is a kite, \Delta PIQ \equiv \Delta RIQ  by side-side-side congruence rule. Hence \angle IRQ (or \angle IRA) = \angle IPQ (or \angle IPB')

Also \angle AIR = \angle B'IP  (same angle).

Hence \Delta AIR, \Delta B'IP  are equiangular. This implies \angle IAR = \angle IB'P

But \angle IB'P = \angle BCC'  (because both of them are subtended by the same segment BC’).

Hence \angle BCC' \left (=\dfrac{\angle C}{2}\right ) = \angle IAR \left (=\dfrac {\angle A}{2} \right)  implying \angle A = \angle C

Screen Shot 2015-12-06 at 11.18.07 PM

Since ABC is isosceles (BA = BC) and BR is the angle bisector, hence it is perpendicular to AC. Thus \angle BRA = \angle IPQ = 90^o  .

Join AB’. \angle AB'P = \angle ACC' = \dfrac{\angle C}{2}  since they are subtended by the same segment AC’ and CC’ is the angle bisector of C.
Similarly \angle IB'P = \angle BCC' = \dfrac{\angle C}{2}  since they are subtended by the same segment BC’ and CC’ is the angle bisector of C.

Clearly \angle IB'P = \angle AB'P = \dfrac{\angle C}{2}  .
Since earlier we found \angle IPB' = \angle APB' = 90^o  ,
therefore \Delta AB'P  is isosceles with AB’=IB’.

Finally AI = IB’ (because \Delta AIR \equiv \Delta B'IP  since earlier we proved them to be equiangular, and we can also show that AR = B’P; How? Clearly \Delta AQP \equiv \Delta B'QR  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 \Delta AIB'  is equilateral, implying \angle AB'I = \angle ACB = 60^o . Since ABC is an isosceles triangle with one angle 60^o  hence it is equilateral.

Chatuspathi:

  • What is this topic: Geometry
  • What are some of the associated concept: Cyclic Quadrilateral
  • Where can learn these topics: Cheenta I.S.I. & C.M.I. course, Cheenta Math Olympiad Program, discuss these topics in the ‘Geometry’ module.
  • Book Suggestions: Challenges and Thrills of Pre-College Mathematics