AoPS Academy
with two altitudes 
and 
.Let 
be the point on the line such that 
.The lines 
and 
intersect at point 
, and 
is the point on segment 
such that 
.Suppose that bisects .Prove that triangle 
is isosceles.
s such that:
such that 
. Prove that
be an acute-angled triangle. Let be a point on the segment 
the incentre of 
The circumcircle of 
meets 
at 
and the circumcircle of 
meets 
at 
If the area of 
and the area of 
are equal, prove that 
be a point on 
such that 
The condition that ![$[PID]=[QID]$](http://latex.artofproblemsolving.com/5/5/9/559a53df3c7705bef486261f81b010aac3a71975.png)
implies that 
is median. Notice that 
Now we cliam that is the foot of on 
![[asy]
size(7cm); pointpen=black; pathpen=black; defaultpen(fontsize(9pt));
pair A=dir(30);
pair B=dir(-147);
pair C=dir(-33);
pair I=incenter(A,B,C);
pair D=foot(I,B,C);
pair E=extension(I,foot(I,A,D),B,C);
pair P=OP(circumcircle(A,B,D),L(B,I,5,5));
pair Q=IP(circumcircle(A,C,D),L(C,I,5,5));
D(circumcircle(A,B,D),dotted);
D(circumcircle(A,C,D),dotted);
D(A--B--C--cycle,purple+linewidth(1.2));
D(A--D);
D(B--P);
D(C--Q);
D(I--E,dashed+red);
D(P--Q,dashed+red);
D(C--E);
D(I--D);
dot("A",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(-75));
dot("$I$",I,dir(90));
dot("$D$",D,dir(-90)*1.5);
dot("$E$",E,dir(E));
dot("$P$",P,dir(110)*1.5);
dot("$Q$",Q,dir(180));
[/asy]](http://latex.artofproblemsolving.com/8/5/1/8517eefbb5c2191e4e1c31b165d2e8af677e138a.png)
We use barycentric coordinate. Set 
and 
where 
and 
Thus![\[\overrightarrow{DA}=(a,-m,m-a),\quad \overrightarrow{EI}=(a,b-t,t-a-b).\]](http://latex.artofproblemsolving.com/3/9/a/39a560143fd9b6f1fa48f6b3d23b5313c1e83afb.png)
By the we get![\[0=a^2(m(a+b)-mt+mb-mt-ab+at)+b^2(at-a^2-ab+ma-a^2)+c^2(ab-at-ma),\]](http://latex.artofproblemsolving.com/9/c/5/9c51e95785127b01132ba5b6b5e14b025d7320af.png)
collect the 
terms we get![\[(2ma-a^2+c^2-b^2)t=am(a+2b)-b(a^2+b^2+2ab-c^2)+m(b^2-c^2).\]](http://latex.artofproblemsolving.com/0/2/c/02c8db82da97400f7eedd4642f5f0bece2af81a5.png)
Substitute 
we get![\[\begin{aligned}
(c^2-b^2)t&=am(b-c)+m(b^2-c^2)-b((a+b)^2-c^2)\\
&=m(b-c)(a+b+c)-b(a+b+c)(a+b-c)\\
&=(a+b+c)(mb-mc-ab-b^2+bc).
\end{aligned}\]](http://latex.artofproblemsolving.com/3/7/6/37691d89ce296d4903c73578b652dccf019222ca.png)
Using 
again we get![\[(c^2-b^2)m=(2m-a)(mb-mc-ab-b^2+bc),\]](http://latex.artofproblemsolving.com/b/d/b/bdb317d09f0cda16b4517661f6cdf66ff16b50a9.png)
which is the following quadratic in 
:![\[2(b-c)m^2-m(c^2+b^2-2bc+3ab-ac)+ab(a+b-c)=0.\]](http://latex.artofproblemsolving.com/0/0/4/004a063af0ac0470ff58252a7ac408df7b52d5ff.png)
Since so we conclude that 
as desired.
and 
so 
are concyclic. Notice that 
is median and 
is isogonal to with respect to 
it follows that 
is harmonic quadrilateral. Therefore 
) and without bary, complex 
be the intersections of 
with 
and 
with 
.
thus 
is cyclic .
then 
since and have the same area and 
and 
are supplementary then 
.
be the intersections of with and with , then is cyclic.
and 
have the same area, will pass through the midpoint 
of 
, then by Ceva's theorem we have 
so 
so 
.
and 
.
since is the -median of 
. And so we are done.
means that is on the median of .
means that should (not yet proved) lie on the Apollonian circle of . should be the 
-point of , which is what we are going to prove.
and are reflections across the line , from 
and 
., , , 
are concyclic by angle chasing, 
. lies on the reflection of circumcircle 
across , which implies that is the -point of as desired.-point is discussed.
such that:
.
are equal means 
besides 
so 
so if 
ie 
but 
implie
therefore 