Post Number:#5 by Ananya Promi » Fri Jun 02, 2017 1:59 pm
Let's draw $RX$ perpendicular on $PK$ which intersects thevcircumcirle of triangle $ABC$ at $M$
$RX$ is parallel to $BC$
So, $BRMC$ is a trapizoid.
So, $BR$=$MC$
Again, $BR$=$AR$ is known as $CR$ in the angle bisector of angle $ACB$
So, $AR$=$MC$ and it makes $ARCM$ cyclic trapizoid
Then $MR$=$b$
$XK$ bisects both $BC$ and also $RM$
So, $RX$=$$\frac{1}{2}b$$
Similarly we can prove $RY$=$$\frac{1}{2}a$$
And draw $RY$ perpendicular on $QL$
Here in the right angled triangle $PKC$,
$PK$=$$\frac{1}{2}atan\frac{1}{2}C$$
Similarly in triangle $LCQ$,
$QL$=$$\frac{1}{2}btan\frac{1}{2}C$$
So, $$[RPK]=\frac{1}{2}PK*RX=\frac{1}{8}abtan\frac{1}{2}C$$
Again, $$[RQL]=\frac{1}{2}QL*RY=\frac{1}{8}abtan\frac{1}{2}C$$