** 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.

** Discussion: **

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’).

Hence implying

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.

Clearly .

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.

## Chatuspathi:

**What is this topic:**Geometry**What are some of the associated concept:**Cyclic Quadrilateral**Where can learn these topics:**Cheenta**Book Suggestions:**Challenges**a**nd Thrills of Pre-College Mathematics

all we need to see is that A,P,R,B’ are concyclic

what if we do not use the information on kite

Information on the kite is critical for the proof. If the the quadrilateral is not a kite, then it is not equilateral.

but i could solve it without the information on kite just as i is the incentre so point R is the point where the incircle meets and its right angle similarly on the left hand side also and just consider the two right triangles RIC and just vertically opposite to it and provet the rest same way……

R is not necessarily the point where incircle meets. It is the point where angle bisector meets the opposite side. You must drop perpendicular from I on the sides to get the points through which incircle pass