MathDB

The circumcircle of triangle

ABC

has centre

O

P

is the midpoint of

\widehat{BAC}

and

QP

is the diameter. Let

I

be the incentre of

\triangle ABC

and let

D

be the intersection of

PI

and

BC

. The circumcircle of

\triangle AID

and the extension of

PA

meet at

F

. The point

E

lies on the line segment

PD

such that

DE=DQ

. Let

R,r

be the radius of the inscribed circle and circumcircle of

\triangle ABC

, respectively. Show that if

\angle AEF=\angle APE

, then

\sin^2\angle BAC=\dfrac{2r}R

Problem Statement