Whoever wrote this... doesn't know what concise means :\
by shiningsunnyday, Apr 11, 2017, 3:46 PM
2004 ISL G2 wrote:
Let 
be a circle and let 
be a line such that and have no common points. Further, let 
be a diameter of the circle ; assume that this diameter is perpendicular to the line , and the point 
is nearer to the line than the point 
. Let 
be an arbitrary point on the circle , different from the points and . Let 
be the point of intersection of the lines 
and . One of the two tangents from the point to the circle touches this circle at a point 
; hereby, we assume that the points and lie in the same halfplane with respect to the line . Denote by 
the point of intersection of the lines 
and . Let the line 
intersect the circle at a point 
, different from . Prove that the reflection of the point in the line lies on the line 
.
be a circle and let 
be a line such that and have no common points. Further, let 
be a diameter of the circle ; assume that this diameter is perpendicular to the line , and the point 
is nearer to the line than the point 
. Let 
be an arbitrary point on the circle , different from the points and . Let 
be the point of intersection of the lines 
and . One of the two tangents from the point to the circle touches this circle at a point 
; hereby, we assume that the points and lie in the same halfplane with respect to the line . Denote by 
the point of intersection of the lines 
and . Let the line 
intersect the circle at a point 
, different from . Prove that the reflection of the point in the line lies on the line 
.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/0/c/6/0c60f19dac196db75f17aa7938bb3bf472112cf8.png)
We show 
is the reflection of 
We'll proceed with straight angle chasing. It suffices to show ![\[\angle{G'AB}=\angle{G'CB}=\angle{BAG}\]](http://latex.artofproblemsolving.com/1/7/9/179c53ddd54aecfb450ca523aa47cd766a93ae02.png)
![\[90-\angle{DCF}=\angle{BAG} \leftrightarrow 90-\angle{BAG}=\angle{PFA}=\angle{DCF} \leftrightarrow \triangle{DCF} \sim \triangle{DFA}\]](http://latex.artofproblemsolving.com/4/3/a/43af76ca0e56f19dc4afd15233ad15fa2fd2c5ef.png)
![\[\frac{DF}{DC}=\frac{AD}{DF} \Rightarrow AD \cdot DC = DF^2.\]](http://latex.artofproblemsolving.com/7/b/5/7b5aad8b4f78989c99e7d393085642ae73997a5c.png)
But by power of a point, 
so it suffices to show 
which follows from a final angle chase: ![\[\angle{DFE}=\angle{PFE}\stackrel{AEPF \text{cyclic}}{=}\angle{BAE}=\angle{DEB},\]](http://latex.artofproblemsolving.com/c/5/1/c51263bb41ac43d7f258345954acd490af1a6a8f.png)
showing 
is isosceles and we're done.Can anyone spot the projective solution (since I found this in a projective chapter in Lemmas).
May 2017
April 2017