[/quote]
,
s.t 
be positive real numbers . Prove that :
and 
be isogonal conjugates with respect to 
.
and 
be their respective pedal triangles. Let![\[ X_1=P_2Q_3\cap P_3Q_2,\quad X_2=P_1Q_3\cap P_3Q_1,\quad X_3=P_1Q_2\cap P_2Q_1 \]](http://latex.artofproblemsolving.com/7/f/e/7fe3ec4f69ec074635b2202daad9371e3bb8c72d.png)
Prove that the points 
,
and 
lie on the line 
.
,
,
be real numbers. Prove that![\[ab + bc + ca + \max\{|a - b|, |b - c|, |c - a|\} \le 1 + \frac{1}{3} (a + b + c)^2.\]](http://latex.artofproblemsolving.com/3/4/9/3499fc076778ca2d6b4ba57c37460d45da166b8c.png)
be integer numbers satisfy that: 
.
?
of an isosceles right-angled triangle 
such 
and 
are marked, such that 
.
.
.
be its circumcircle, 
its orthocenter, and 
the foot of the altitude from 
.
be the midpoint of 
.
such that 
and let 
.
,
,
and 
are tangent to each other.
be positive real numbers. Find the minimum value of![\[ T = (x+y+z+t)^2.\left[\dfrac{1}{x(y+z+t)}+\dfrac{1}{y(z+t+x)}+\dfrac{1}{z(t+x+y)}+\dfrac{1}{t(x+y+z)}\right] \]](http://latex.artofproblemsolving.com/c/a/c/cac0967ebdf30c1f79a1ec3e468cc035402333a7.png)
,
,
,
be the diametrically opposite point of 
and 
intersect the incircle again at 
and 
,
,
,
are concurrent, i.e., the lines intersect at the same point.
and 
be two similar right-angled triangles (in the same plane) with 
such that 
.
,
.
is written on the blackboard. Every time, Carmela can erase the number 
on the black board and replace it with a new number 
,
is a perfect square. Prove or disprove that all positive integers 
can be written exactly once on the blackboard.
,
such that 
lie on 
be its center. Let 
be some point lying on the circumcircle of 
.
is tangent to 
is tangent to 
and 
,
meet the circumcircle of 
and 
.
,
.
and 
exist on 
and 
are tangent to 
,
,
,
.
at the orthocenter of 
,
be a point lying on 
such that 
.
.
be its incircle, and let 
be the midpoint of minor arc 
on the circumcircle of 
at points 
,
and 
.
and 
denote the 
,
,
and 
,
and 
.
and 
are tangent to each other.
and 
be circles tangent to 
meets 
.
similarly.
and 
.
and 
.
Y by NO_SQUARES, Mogmog8, ImSh95, jmiao, Rounak_iitr, HoRI_DA_GRe8
Let 
be an acute scalene triangle with orthocenter 
. Line 
intersects 
at 
and line 
intersects 
at 
. Let 
be the foot of the perpendicular from to the line through 
parallel to 
. Point 
lies on line 
such that 
is parallel to , and point 
lies on line 
such that 
is parallel to . Prove that points 
, 
, , are concyclic.
Proposed by Luke Robitaille
be an acute scalene triangle with orthocenter 
. Line 
intersects 
at 
and line 
intersects 
at 
. Let 
be the foot of the perpendicular from to the line through 
parallel to 
. Point 
lies on line 
such that 
is parallel to , and point 
lies on line 
such that 
is parallel to . Prove that points 
, 
, , are concyclic.Proposed by Luke Robitaille
.
are concyclic. Now 
implies that 
is an isosceles trapezoid.
is cyclic.
.
.
is cyclic. Now 
.
are collinear.
.
they subtend equal angles. Thus 
are concyclic and we are done.
be the 
point
is a parallelogram and that 
are collinear.
are concyclic, and since 
,
point that 
and 
are cyclic quadrilaterals.
lies on 
and a similarly , a conclusion will follow that 
lies on 
.
.
d
and 
is cyclic, or in other words 
so 
is cyclic, as desired
.
passes through the midpoint of 
and so 
.
is a trapezoid. Let 
.
.
as desired 
.
and so 
and 
is cyclic
and 
,
and 
is an isosceles trapezoid. So, 
.
,
and 
so 
and 
.
.
as 
.
.
,
and so, 
is cyclic.
So, 
.
.
.
Hence, 
tangent to 
and 
are cyclic.
and 
tangebt to 
so 
.
,
then 
wich gives the result
are collinear
are cycilc
and 
cyclic we get 
,
and 
.
and 
are similar isosceles trapeziums and 
because they are the corresponding figures in the trapeziums, as a result 
are collinear. 
so 
and similarly 
so 
,
so 
.
which means 
and equal angles give 
are cyclic. Because 
,
is the radical axis of these two cyclic quadrilaterals and power of point from 
which finishes our problem.
meet 
.
and then it is easy trig stuff after noting 
,
.
.
so that the diagram looks like this...
.
so 
and then it will just be a matter of doing some trig manipulations...
and as we already know 
we just gotta find the angles of that triangle to solve this problem by law of sines.
is clear due to parallel lines.
,
,
(due to tangent) and 
(due to cyclic pentagon 
)
.
,
(This is important to note, it is 
factor)
(bcoz 
)
by 
everywhere.
,
Similarly, 
.
,
and 
(
hits 
,
.
.
and ![[asy]
import geometry;
defaultpen(fontsize(13pt));
size(10cm);
pair A = dir(120);
pair B = dir(210);
pair C = dir(330);
pair H = orthocenter(A,B,C);
pair E = foot(B,A,C);
pair F = foot(C,A,B);
pair M = (B+C)/2;
pair X = 2*extension(bisectorpoint(E,F), (E+F)/2, A,foot(A,bisectorpoint(E,F), (E+F)/2)) - A;
pair B1 = intersectionpoints(line(X,F), circumcircle(B,M,F))[1];
pair C1 = intersectionpoints(line(X,E), circumcircle(E,M,C))[0];
draw(A--B--C--A, purple);
draw(circumcircle(M,E,C), orange);
draw(circumcircle(M,F,B), orange);
draw(C--F, purple);
draw(B--E, purple);
draw(C--C1, purple);
draw(M--E, purple);
draw(X--C1, purple + dashed);
draw(X--B1, purple + dashed);
draw(B1--C1, purple + dashed);
draw(E--F, purple);
draw(circumcircle(X,A,E), red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$H$", H, S);
dot("$E$", E, 2*dir(75));
dot("$F$", F, dir(F));
dot("$M$", M, dir(M));
dot("$X$", X, dir(X));
dot("$B_1$", B1, dir(B1));
dot("$C_1$", C1, dir(C1));
[/asy]](http://latex.artofproblemsolving.com/f/1/2/f12bde32b41ffc2d3d5f299f64fba0fc2bce1c0a.png)
and 
.
are collinear. This can be shown easily by an angle chase.
and thus the claim holds true. 
are collinear. Note that![\[\measuredangle EC_1M = \measuredangle ECM = \measuredangle EFA = \measuredangle XEF\]](http://latex.artofproblemsolving.com/5/7/1/571fa5480cd0eef713bb36932ac29a216a15876b.png)
which shows that 
.
.
and the solution is complete. 
and let 
Also denote by 
a directed angle.
so 
is cyclic. Moreover, 
so 
and 
are isosceles.
since 
We also know that by sum of angles 
However, since 
we get 
so 
On top of that, we know that 
since 
and 
is cyclic. Therefore, 
However. we know that 
so 
is isosceles. Hence 
so 
This implies that 
is cyclic. Similarly, 
is also cyclic.
so 
are collinear. Finally, 
from our parallel lines, so 
,
,
.
and 
,
is cyclic. Let 
is a parallelogram, and thus 
so 
be a point on 
such that 
.
gives 
parallel to the tangent to 
and 
)
which proves the claim.
which is the tangent to 
and 
.
and 
,
is the perpendicular bisector of 
,
,
.
are similar since 
and 
.
.
and 
,![\[\frac{YE}{ZF}=\frac{YF}{ZF}=\frac{\sin (180^\circ-\angle YZA)}{\sin \angle FYZ}=\frac{\sin \angle EFH}{\sin \angle FEH}=\frac{HE}{HF}=\frac{CE}{BF},\]](http://latex.artofproblemsolving.com/9/c/7/9c79541ba724fb2fb58735ab76df7e6278acd41c.png)
where the last equality follows from 
.
and 
,![\[\dfrac{XF}{FB_1}\cdot \dfrac{FB}{ZF}=\dfrac{XE}{EC_1}\cdot \dfrac{EC}{YE}\iff \dfrac{XF}{FB_1}=\dfrac{XE}{EC_1}.\]](http://latex.artofproblemsolving.com/a/f/0/af01fe52166223ce05faa9851c776722f8e07fb7.png)
We then obtain that 
and that 
.
,
be such that 
is a parallelogram, i.e. 
are collinear. Let 
is cyclic, so by PoP at 
and 
are also cyclic (this is well-known, actually). We claim that 
and 
;
,
,
,
is cyclic and similarly 
is cyclic, as wanted.
is a cyclic isosceles trapezoid.
.
.
is the radical axis of those two circles. To finish, let 
.
,
,
!
,
,
.
where 
(near 
as the intersections of 
and 
be 
and 
,
and 
.
,
.
we have that
hence
Since 
,
,
and 
are isosceles, hence 
and 
,
and 
.
is cyclic, so 
.![\[ TB \cdot TB_1 = TC \cdot TC_1,\]](http://latex.artofproblemsolving.com/1/3/8/13856ddb2cafa6f11deafe499b8a211738e1f4cd.png)
so 
,
,
is cyclic. Similarly, 
is cyclic. However, it is well known that 
and 
meet at the 
,
,
.
and as 
.![\[\angle FLE = 2 \angle FXH\]](http://latex.artofproblemsolving.com/e/e/5/ee522be107d47a19a2e8d383bfcac8769de10802.png)
are collinear.
and 
is cyclic. Since 
is cyclic too ![\[\angle BHP = \angle BCP = \angle ABC = \angle AEF \stackrel{XF = AE}{=} \angle XFE = \angle XHE \square \]](http://latex.artofproblemsolving.com/e/a/3/ea3c4ab720856184b80e64631bec0848d4085187.png)
.![\[ 2 \angle KXL = 2\angle FXH = \angle FLE = \angle XKL + \angle KXL \]](http://latex.artofproblemsolving.com/9/8/5/985088752d56a84a609e7c308c12e8c6924152c4.png)
then 
and ![\[\angle B_{1}PX = \angle XKL = \angle KXL = \angle FXH = \angle B_{1}XP \]](http://latex.artofproblemsolving.com/a/b/1/ab167e92305901f59a148ea0bf3f1f9906fb41bf.png)
so 
.
therefore 
is a deltoid with 
,
.![\[\angle B_{1}BC = \angle BCA = \angle AFE = \angle XEF = \angle XC_{1}B_{1} = \angle BC_{1}P = \angle B_{1}C_{1}C \]](http://latex.artofproblemsolving.com/6/c/e/6ce569c68563ecfb19e9fe36d5bd1fdca47079f8.png)
thereby 
and 
are concyclic ![[asy]
import graph;
size(10cm);
pair A = (-2.72796,12.27769);
pair B = (-6.14859,2.36593);
pair C = (5.56774,3.20089);
pair E = (0.65108,8.58048);
pair F = (-4.64420,6.72512);
pair H = (-2.27397,5.90713);
pair X = (-4.31165,11.72279);
pair B_1 = (-5.01667,1.12743);
pair C_1 = (6.20965,5.06090);
pair P = (2.14711,-6.71086);
pair K = (-5.61156,15.43279);
pair L = (-4.16976,13.85524);
pen zzttqq = rgb(0.6,0.2,0);
pen ccqqqq = rgb(0.8,0,0);
pen fuqqzz = rgb(0.95686,0,0.6);
pen qqwuqq = rgb(0,0.39215,0);
draw((-4.12839,11.19976)--(-3.60536,11.38302)--(-3.78862,11.90605)--X--cycle, linewidth(2) + ccqqqq);
draw((-0.55924,2.68922)--(-0.74250,3.21225)--(-1.26553,3.02899)--(-1.08227,2.50596)--cycle, linewidth(2) + ccqqqq);
draw(arc(B,0.78376,-47.57435,4.07625)--B--cycle, linewidth(2) + qqwuqq);
draw(arc(C_1,0.78376,-160.69061,-109.03999)--C_1--cycle, linewidth(2) + qqwuqq);
draw(arc(C,0.78376,132.42564,184.07625)--C--cycle, linewidth(2) + qqwuqq);
draw(arc(E,0.78376,147.65877,199.30938)--E--cycle, linewidth(2) + qqwuqq);
draw(arc(C_1,0.78376,147.65877,199.30938)--C_1--cycle, linewidth(2) + qqwuqq);
draw(arc(F,0.78376,19.30938,70.96000)--F--cycle, linewidth(2) + qqwuqq);
draw(arc(K,0.78376,-70.69061,-47.57435)--K--cycle, linewidth(2) + qqwuqq);
draw(arc(X,0.78376,86.19313,109.30938)--X--cycle, linewidth(2) + qqwuqq);
draw(arc(L,0.78376,-93.80686,-70.69061)--L--cycle, linewidth(2) + qqwuqq);
draw((-1.47352,7.83606)--(-1.65678,8.35909)--(-2.17981,8.17583)--(-1.99655,7.65280)--cycle, linewidth(2) + ccqqqq);
draw(arc(P,0.78376,109.30938,132.42564)--P--cycle, linewidth(2) + qqwuqq);
draw(arc(E,0.78376,-160.69061,-137.57435)--E--cycle, linewidth(2) + qqwuqq);
draw(B--A, linewidth(2));
draw(B--C, linewidth(2));
draw(F--C, linewidth(2));
draw(B--E, linewidth(2));
draw(F--E, linewidth(2));
draw(P--X, linewidth(2) + linetype("4 4"));
draw(P--C_1, linewidth(2) + zzttqq);
draw(X--C_1, linewidth(2) + zzttqq);
draw(X--A, linewidth(2));
draw(circle((-0.54467,6.35111), 6.87645), linewidth(2) + dotted);
draw(circle((-2.50096,9.09241), 3.19335), linewidth(2) + ccqqqq);
draw(X--K, linewidth(2) + linetype("4 4"));
draw(K--A, linewidth(2));
draw(L--X, linewidth(2));
draw(E--A, linewidth(2));
draw(E--C, linewidth(2));
draw(B_1--F, linewidth(2) + fuqqzz);
draw(F--X, linewidth(2) + fuqqzz);
draw(B_1--C_1, linewidth(2));
draw(B--B_1, linewidth(2));
draw(B_1--P, linewidth(2) + fuqqzz);
draw((-1.99655,7.65280)--L, linewidth(2) + linetype("4 4"));
dot("$A$", A, dir((20.446, 72.656)));
dot("$B$", B, dir((9.813, 26.646)));
dot("$C$", C, dir((79.145, -38.562)));
dot("$E$", E, dir((11.723, 21.755)));
dot("$F$", F, dir((-195.489, -4.327)));
dot("$H$", H, dir((11.623, 19.996)));
dot("$X$", X, dir((-111.179, 41.931)));
dot("$B_1$", B_1, dir((-137.342, -126.434)));
dot("$C_1$", C_1, dir((88.105, 21.017)));
dot("$P$", P, dir((-169.231, -84.572)));
dot("$K$", K, dir((10.974, 21.014)));
dot("$L$", L, dir((10.485, 22.015)));
[/asy]](http://latex.artofproblemsolving.com/f/8/4/f84e9a5e7ae5dbc898c342810bb54ea33f81f5de.png)
.
.
,
.
is cyclic by Reim after reflecting over the midpoint of 
.
,
,
.
.
,
We get
Therefore, 
is a parallelogram, so 
and 
.
.
since 
is cyclic 
)
are collinear. Further, this concyclicity also gives
Similarly, we find that 
.
and 
,
.
and 
,
.
hence 
Then we observe that
Thus 
Now we find the coordinate of 
,
Since 
,
Setting these two equations for 
equal, we have
Similarly,
Then we find the vectors
Thus
which is real. 
,
is a parallelogram.
,
collinear. But ![\[\measuredangle B_1XA'=\measuredangle FEH=\measuredangle FCB=\measuredangle XA'B_1,\]](http://latex.artofproblemsolving.com/3/0/2/30270f63a1df4a764c0f59d33bf367de196f70fb.png)
implying 
.
,
.
and 
are antiparallel in 
,
is indeed cyclic.![[asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=darkgreen; pen pri2=heavygreen; pen sec=heavycyan; pen tri=lightblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,H,EE,F,X,B1,C1,Ap,M,P; A=dir(110); B=dir(215); C=dir(325); H=A+B+C; EE=foot(B,A,C); F=foot(C,A,B); X=foot(H,A,A+EE-F); B1=extension(X,F,B,B+C-A); C1=extension(X,EE,C,C+B-A); Ap=B+C-A; M=(B+C)/2; P=reflect(circumcenter(B,F,M),circumcenter(C,EE,M))*M; draw(A--Ap,tri+dashed); filldraw(circumcircle(B,F,M),tfil,tri); filldraw(circumcircle(C,EE,M),tfil,tri); draw(B1--C1,tri+dashed); draw(B1--X--C1,sec); filldraw(circumcircle(A,EE,F),sfil,sec+dashed); draw(X--Ap,pri2+dashed); draw(B--EE,pri2+linewidth(.4)); draw(C--F,pri2+linewidth(.4)); draw(B--Ap--C1,pri); fill(B--Ap--C--cycle,fil); filldraw(A--B--C--cycle,fil,pri); dot("\(A\)",A,N); dot("\(B\)",B,B); dot("\(C\)",C,C); dot("\(M\)",M,S); dot("\(E\)",EE,dir(60)); dot("\(F\)",F,dir(200)); dot(P); dot("\(A'\)",Ap,S); dot("\(H\)",H,S); dot("\(X\)",X,NW); dot("\(B_1\)",B1,SW); dot("\(C_1\)",C1,E); [/asy]](http://latex.artofproblemsolving.com/9/4/d/94d3e94e3e4bab5be40f334b1cebdaed7ea16400.png)
be the midpoint of 
is an isosceles trapezoid and 
,![\[\measuredangle B_1FM=\measuredangle XFM=\measuredangle MEA=\measuredangle ECM=\measuredangle B_1BM,\]](http://latex.artofproblemsolving.com/f/6/f/f6f42f73712406a0b6010b5928e513cb255b71bd.png)
so 
.
.
,
is the radical axis of 
and 
.
as needed.
be the perpendicular bisector of 
)
is the reflection of 
be the midpoint of 
.
and 
are reflections in 
,
,
.
.
,
is a parallelogram 
is cyclic
and 
also 
similarly 
are concyclic.
.
.
and 
.
are cyclic.
because 
is cyclic. 
and 
,
is the reflection of 
,
is a parallelogram, 
,
.
.
be the inverted version of 
; hence, 
.
.
=
are collinear.
.
are concyclic. Similarly, 
are concyclic.
lie on a circle.
and we also have 
.
This implies 
where 
is cyclic as well.Its well known that they meet at the Humpty point 
it's well known that 
,
and 
are cyclic. There is a point 
.
and 
,
.
and 
,
be the reflection of 
are cyclic
.
,
is isosceles with 
.
so 
from cyclic quadrilateral 
and 
is line 
,
.
and ![\[\measuredangle CC_1E = \measuredangle CEC_1 + \measuredangle C_1CE = \measuredangle AEX + \measuredangle BAC = 2\measuredangle BCA\]](http://latex.artofproblemsolving.com/8/c/2/8c203e2f00e62df71df0f2cc9a7cb213f45ff485.png)
A similar angle chase shows that 
.
and 
,![\[XE = \frac{AE \sin \angle XAE }{\sin \angle AXE}\]](http://latex.artofproblemsolving.com/4/e/8/4e8cfdb538cb1165aa01eeeff69bd63c716621bc.png)
and![\[EC_1 = \frac{EC \sin \angle C_1CE}{\sin \angle CC_1E}\]](http://latex.artofproblemsolving.com/f/4/5/f45a866d8a004bcadc4dc42e343fceddc0626378.png)
So,![\[\frac{XE}{CC_1}= \frac{AE}{EC} \cdot \frac{\sin B \sin 2C}{\sin C \sin A}\]](http://latex.artofproblemsolving.com/5/8/5/5851308abd563809baf03e3d09f457d971966b0f.png)
A similar calculation also yeilds,![\[\frac{XF}{FB_1} = \frac{AF}{FB} \cdot \frac{\sin C \sin 2B}{\sin B \sin A}\]](http://latex.artofproblemsolving.com/c/7/3/c73b7cd85f301818237256617ab30bb9abe7dc8e.png)
Thus,
which indeed implies that 
as claimed.
.![\[\measuredangle A'HB = \measuredangle A'CB = \measuredangle ABC = \measuredangle FEA = \measuredangle XAE = \measuredangle XHE\]](http://latex.artofproblemsolving.com/5/1/c/51c5605fa416edae947b3a40343d97f39afd015a.png)
which implies the claim.
and 
.![\[\measuredangle B_1TH + \measuredangle HTC_1 = \measuredangle HBB_1 + \measuredangle HCC_1 = \pi\]](http://latex.artofproblemsolving.com/2/1/b/21bddb103ebca72c7fd324103962d7864acee201.png)
so points 
,
.![$HT ]perp B_1C_1$](http://latex.artofproblemsolving.com/0/b/9/0b9240916cbf803bb50ed79b887bbed9c58c0a11.png)
as well this implies that points ![\[AB_1\cdot AB = AT \cdot AH = AC_1 \cdot AC \]](http://latex.artofproblemsolving.com/6/1/a/61addbe138237f1bd3b6f45f7be667cdf0d2fa82.png)
which implies that points 
and 
intersect at ![\[\frac{\operatorname{dist}(X,\overline{A'B})}{\operatorname{dist}(F,\overline{A'B})}=\frac{\operatorname{dist}(X,\overline{A'C})}{\operatorname{dist}(E,\overline{A'C})}.\]](http://latex.artofproblemsolving.com/b/7/f/b7f3bf70a8d5ab0e7343dec4553fe91972719cea.png)
By angle chasing, we know ![\[\frac{\operatorname{dist}(X,\overline{A'B})}{\operatorname{dist}(H,\overline{A'B})}=\frac{\operatorname{dist}(X,\overline{A'C})}{\operatorname{dist}(H,\overline{A'C})}\]](http://latex.artofproblemsolving.com/9/5/4/95458e39a5a37ce2457eced45f1af045dcf515c8.png)
From 
,
,
,![\[\frac{\operatorname{dist}(H,\overline{A'B})}{\operatorname{dist}(F,\overline{A'B})}=\frac{\operatorname{dist}(H,\overline{A'C})}{\operatorname{dist}(E,\overline{A'C})}.\]](http://latex.artofproblemsolving.com/0/8/9/08910518abdfa352d937e615c6ff20148e909009.png)
Multiplying these two equations gives us the desired result. 
intersect 
intersect 
at 
is an isosceles trapezoid.
![\[\angle HXA=90^\circ=\angle HEA=\angle HFA\]](http://latex.artofproblemsolving.com/2/e/8/2e83848839cdc88e3f76bb765fd9c4d0934abe07.png)
which proves that 
and 
are isosceles triangles.
,
which is symmetric, so it is also equal to 
.
,![\[\frac{XF}{FQ}\cdot \frac{AF}{FB}=\frac{AE}{EC}\cdot \frac{XE}{EP}\]](http://latex.artofproblemsolving.com/1/a/f/1afb848443736b9a882d48c47599edb9ff694c66.png)
The left hand side is![\[\frac{XF}{FQ}\cdot \frac{AF}{FB}=\frac{XF}{FQ}\cdot \frac{QF}{FB_1}=\frac{XF}{FB_1}\]](http://latex.artofproblemsolving.com/7/6/7/767d46ef00f158a8ab152f2858950c15dd79dd22.png)
and the right hand side is![\[\frac{XE}{EP}\cdot \frac{AE}{EC}=\frac{XE}{EP}\cdot \frac{PE}{EC_1}=\frac{XE}{EC_1}\]](http://latex.artofproblemsolving.com/6/1/6/616aecb271fbf68542e4811fb1bdba42d60513b4.png)
and the two are equal, which proves the claim.
.![\[\measuredangle XC_1A'=\measuredangle XC_1C=\measuredangle XPF=2\measuredangle XEF=2\measuredangle XC_1B_1\]](http://latex.artofproblemsolving.com/7/a/3/7a3d99d8d2ead2a28a9643397df55ecad8c179b5.png)
so 
and similarly bisects 
.![\[\triangle A'B_1C_1\sim\triangle XB_1C_1\sim\triangle XFE\sim\triangle AEF\sim\triangle ABC\sim\triangle A'CB\]](http://latex.artofproblemsolving.com/c/a/7/ca78c039041bdf53698026154f890b6d2d09b6f9.png)
as desired.
which rearranges to 
so by power of a point, we are done.
and 
respectively. Let 
and 
intersect at 
is a parallelogram.
,
.
Clearly, 
cyclic, which readily implies 
,
Therefore 
is isosceles and likewise, 
is isosceles. This implies that 
is simply the reflection of 
over 
.
and 
,
and 
.
,
concyclic
concyclic 
is isosceles trapezoid
collinear
.
concyclic.
