1998 JBMO Problems/Problem 2
Let
Let angle =
Applying cosine rule to triangle we get:
Substituting we get:
From above,
Thus,
So, of triangle
=
Let be the altitude of triangle DAC from A.
So
This implies .
Since is a cyclic quadrilateral with
, traingle
is congruent to
.
Similarly
is a cyclic quadrilateral and traingle
is congruent to
.
So of triangle
+
of triangle
=
of Triangle
.
Thus
of pentagon
=
of
+
of
+
of
=