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be an acute triangle with 
, and let 
be the projection from 
onto 
. Let 
be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
sastifying 
and the angle bisector 
of a right triangle (
) intersect at the point 
. Find the area of the triangle if 
.
![$\frac{[AOD]}{[AOC]}=\frac{OD}{CO}=\frac{5}{9} \implies \boxed{[AOD]=\frac{5}{9} [AOC]}$](http://latex.artofproblemsolving.com/8/3/6/8364d10155a762428da98d227574342e88bfbd1d.png)
![$[ABC]=x$](http://latex.artofproblemsolving.com/d/d/0/dd0f2f7518cd7231450b71bc450d28651de3df73.png)
, Let 
, Therefore, ![$$[ADC]=\frac{bx}{a+b} ; [AOC]=\frac{bx}{a+2b} ; [AOD]=\frac{5}{9}(\frac{bx}{a+2b})$$](http://latex.artofproblemsolving.com/c/6/1/c6124d3021d371c8e5af92504ae1bda49736d38c.png)
Since, ![$$[ADC]=[AOD]+[AOC] \implies \frac{bx}{a+b}=\frac{bx}{a+2b}+\frac{5}{9}(\frac{bx}{a+2b}) \implies 9a+18b=14a+14b \implies \boxed{a=\frac{4b}{5}}$$](http://latex.artofproblemsolving.com/5/9/1/5916b77228e5cf5c99587606e8cbeaa3a4a458ab.png)
Sine Rule immediately follows that, 
and 
![$$[AOC]=\frac{bx}{a+2b}=\frac{bx}{\tfrac{4b}{5}+2b}=\frac{5x}{14} =\frac{1}{2}9b\sin \frac{C}{2} \implies 5x=63b\sin \frac{C}{2}=\frac{63b}{\sqrt{10}} \implies \boxed{x=\frac{63b}{5\sqrt{10}}}$$](http://latex.artofproblemsolving.com/e/4/4/e44d66dcfb8a0391558b8ffadcfa23ccbce714de.png)
