be an acute triangle inscribed in a circle 
.
and 
intersect at 
.
be a point on the minor arc 
. The lines 
and 
intersect the lines 
and 
,
passes through the midpoint of segment 
.
be real numbers such that
Prove
Y by GeoKing, OronSH, SerdarBozdag
Let 
be a triangle with points 
lie on the perpendicular bisector of 
such that 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
to the circumcircle of 
intersects at point 
Prove that: 
are parallel.
Proposed by Paramizo Dicrominique
be a triangle with points 
lie on the perpendicular bisector of 
such that 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
to the circumcircle of 
intersects at point 
Prove that: 
are parallel.Proposed by Paramizo Dicrominique
This post has been edited 1 time. Last edited by Noob_at_math_69_level, Dec 18, 2023, 7:27 PM
be the foot from 
to 
Since 
and 
we see that 
and 
Furthermore, we simply notice that the problem is equivalent to showing that 
is parallel to the 
-
by PoP. Denote the image of a point 
under this inversion by 
We check that 
is arbitrary, 
are points on 
respectively, with 
cyclic, and 
is the second intersection of 
and 
as shown before. We see 
and the 
inverts to the 
(a fact which may be easily verified with the inversion distance formula). Thus, letting 
be the midpoint of 
we now wish to prove that 
is tangent to 
with directed angles. We wish to show this. First, from the definition of 
Since the spiral similarity taking 
takes 
we get that 
Thus we have 
is isosceles, letting 
intersect 
we get 
again with directed angles, so 
Thus the spiral similarity at 
to 
so we have 
finishing.
to be second intersection of 
with 
.
.
be the second intersection 
)
and 
.
are collinear.
while 
.
is parallel to 
.
.
and suppose 
cyclic. Let 
denote the second intersection of 
and 
and let 
denote the second intersection of 
and 
.
.
.
.
,
.
are midpoints of 
respectively, we therefore get ![\[\triangle IBF \sim \triangle ICE \sim \triangle IDG\]](http://latex.artofproblemsolving.com/8/9/2/892c246ba24f486dac25a21c701a3dd024fec95a.png)
To finish, we have ![\[\measuredangle IDG = \measuredangle IBF = \measuredangle IBU = \measuredangle IVU = \measuredangle IVD\]](http://latex.artofproblemsolving.com/3/7/5/3753d541363d56566fd1db8fd1684eb95c79eacd.png)
which implies that line 
is tangent to 
,
is fixed as 
is projective from 
.
and similarly 
.
at most 
.
is at most 
is at most 
to get 
is projective from the perp bisec to itself. Thus 
and so 
.
has degree at most 
.
cases that the intersection is a point at infinity. For the cases, we have.
as 
is trivial.
follows by using harmonics.
follows by Simson's Theorem.
follows by a homothety at the centroid mapping the medial triangle to 
,
.
,
,
.
and 
with 
respectively. We claim that 
and 
.
.
Then, by inverting at 
,
,
and 
.
,
.
,
,
to 
.
,
,
,
is isosceles, and so is 
.
intersect 
,
,
,
,
,
as 
;
.
be the tangent at 
because
Now we simply have ![\[-1 = (BC; UV) \stackrel{A}{=} (ZY; LK) = V(ZY; LK) = D(FE; DS).\]](http://latex.artofproblemsolving.com/e/e/c/eecb4eb876174cd2b18d1c924c2f6c82be86060f.png)
Thus it follows that 
,
,
which is on 
and 
.
.
is cyclic. By radical axis theorem on 
,
and 
,
.
,
.
.
which gives 
,
.
.
and 
with 
and 
.![\[\measuredangle BXV=\measuredangle BUV=\measuredangle BUD=\measuredangle BFD\]](http://latex.artofproblemsolving.com/f/b/e/fbe5ed1330a4fa92cff87be2ca05b1ad42c9501e.png)
and the analogous.
.
and we will prove it by showing that![\[\frac{\sin(\angle SDE)}{\sin(\angle SDF)}=\frac{\sin(\angle AVY)}{\sin(\angle AVX)}\]](http://latex.artofproblemsolving.com/e/c/f/ecf335e7b8959d0913dcfac2e4435d2e80cfddae.png)
Indeed, both are equal to 
.
.
since 
Noting that 
and 
finishes.
.
we have![\[\frac{\sin(\angle AVY)}{\sin(\angle AVX)} = \frac{AY}{AX}=\frac{AB}{AC}\]](http://latex.artofproblemsolving.com/b/f/d/bfd81421cdae6d50995f4993f3d3b97e162f8b9b.png)
as desired.
