be the circumcircle of a scalene triangle 
be a circle internally tangent to 
.
touch 
and 
,
.
touch 
and 
,
and 
concur on the bisector of 
.
,
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
and 
are the midpoints of sides 
and 
.
.
,
,
of a hexagon 
inscribed in a circle 
intersect at a point 
,
.
be the radius of the circle tangent to segments 
and 
and internally tangent to 
,
,
,
,
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
Y by Adventure10, Mango247
The incircle of a triangle 
touches the sides 
, 
, 
at the points 
, 
, 
respectively. Let the line 
intersect this incircle of triangle at a point 
(apart from ). Assume that this point is the midpoint of the segment , this means, 
. Let the line 
meet the incircle of triangle at a point 
(apart from ), and let the line 
meet the incircle of triangle at a point 
(apart from ). Show that 
.
touches the sides 
, 
, 
at the points 
, 
, 
respectively. Let the line 
intersect this incircle of triangle at a point 
(apart from ). Assume that this point is the midpoint of the segment , this means, 
. Let the line 
meet the incircle of triangle at a point 
(apart from ), and let the line 
meet the incircle of triangle at a point 
(apart from ). Show that 
.
Power of the vertex A to the incircle (I) is 
Since 
we get 
BX is the B-median of the triangle 
Thus
In exactly the same way, we can show that 
As a result, EFYZ is a cyclic trapezoid with bases 
i.e., isosceles and with equal diagonals, hence 
,
.
,
.
,
should be parallel to 
should be parallal to 
where 
and 
.
,
,
and 
.![\[AX AD=AE AE\]](http://latex.artofproblemsolving.com/9/3/c/93caff3f2af3b6860646f8be2043925780283396.png)
![\[AD = (s-a)\sqrt{2}\]](http://latex.artofproblemsolving.com/5/a/9/5a909b8347358201ac5d614a89b494251e7d318e.png)
From triangle ADC we get![\[\frac{\sin \varphi}{\sin \gamma}= \frac{CD}{AD}\]](http://latex.artofproblemsolving.com/4/1/7/417e7cbf461aad0d1c90986ba0144a5ce4bf6055.png)
![\[\sin \varphi= \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\gamma}{2}\]](http://latex.artofproblemsolving.com/3/3/7/337d63702d62161786276a2f0a5dc02530d7d8ed.png)
![\[\sin \theta = \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\beta}{2}\]](http://latex.artofproblemsolving.com/7/6/5/7657af71cb7cc74df44169e0bf4a0ef530ade9f4.png)
In triangle AXB: ![\[\frac{\sin \theta}{\sin \xi}= \frac{BX}{AX}\]](http://latex.artofproblemsolving.com/7/b/b/7bbd38c28039a075bc6d8a9f5792db2442327b7a.png)
From triangle BFY: ![\[\frac{\sin \mu}{\sin \xi}= \frac{BY}{FY}\]](http://latex.artofproblemsolving.com/c/7/f/c7f11ff3938970fd8dedb1111b79dae3f6b676d5.png)
We have: ![\[BX BY = BD^{2}\]](http://latex.artofproblemsolving.com/a/0/7/a0794d8e643604053100d0467a00a01ee871b6da.png)
![\[\sin \xi = \sqrt{2}\sin \mu \sin \frac{\beta}{2}\]](http://latex.artofproblemsolving.com/a/2/5/a25e513203a495ff3ed1360f16ea0865f784c43c.png)
Also we have ![\[\sin \nu = \sqrt{2}\sin \eta \sin \frac{\gamma}{2}\]](http://latex.artofproblemsolving.com/6/7/0/6701c3c33033da33415e5decaa72b50d09748f6a.png)
From Ceva' s theorem we have![\[\frac{\sin \varphi}{\sin\theta}\frac{\sin \xi}{\sin \nu}=\frac{\sin(\beta-\xi)}{\sin(\gamma-\nu)}= \frac{CX}{BX}\]](http://latex.artofproblemsolving.com/8/6/9/869d298f4e7f0d9b79dc9b9c7d96ce98552e0abf.png)
but I don't know how to
and 
then using Pascal's theorem on hexagon 
we obtain 
and 
are collinear, and because 
,
.
,
making 
an iscoceles trapezium. So the diagonals are congruent hence 
be the midpoint of 
.
and that 
is the 
s
,![\[
\angle BFY=\angle FXY=\angle MXD=\angle FAD
\]](http://latex.artofproblemsolving.com/8/0/f/80fd56fe2f42e13cf49eea31a510441dce9a6792.png)
meaning that 
