[/quote]
,
,
to the sides 
are 
,
is the orthocenter. 
and 
intersect at 
from 
.
are colinear
and 
,
as the greatest exponent 
such that 
divides 
.
and infinitely many 
such that 
.
prime and 
satisfying the equation: 
be positive real numbers with 
.
and 
satisfy 
,
is an even number.
,
,
for which the equality holds.
.
,
.
,
lines intersect at point 
,
.![\[(a-b)^2 + a + b \geq 2\]](http://latex.artofproblemsolving.com/c/f/a/cfa01a5b54a72afe6746f547390e65dbbb82ea64.png)
and 
we define 
as the set of all products of an element of 
and 
then 
We call a set 
of positive integers good if there do not exist sets 
of positive integers, each with at least two elements and such that the sets 
and 
is good.
,
and 
are integers.)
Y by OronSH, peace09, Rounak_iitr, ehuseyinyigit, Funcshun840
Let 
be an acute-angled triangle with circumcircle 
. A circle 
is internally tangent to at 
and also tangent to 
at 
. Let 
and 
intersect at 
and 
respectively. Let 
and 
be points on line such that 
is the midpoint of 
and 
is the midpoint of 
. Lines 
and 
meet at 
and intersect again at 
and 
respectively. The ray 
meets the circumcircle of triangle 
again at 
.
Prove that
.
Kian Moshiri, United Kingdom
be an acute-angled triangle with circumcircle 
. A circle 
is internally tangent to at 
and also tangent to 
at 
. Let 
and 
intersect at 
and 
respectively. Let 
and 
be points on line such that 
is the midpoint of 
and 
is the midpoint of 
. Lines 
and 
meet at 
and intersect again at 
and 
respectively. The ray 
meets the circumcircle of triangle 
again at 
.Prove that
.Kian Moshiri, United Kingdom
This post has been edited 2 times. Last edited by MarkBcc168, Jul 18, 2024, 8:50 PM
![[asy]
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defaultpen(fontsize(8));
pair A = dir(120);
pair B = dir(200);
pair C = dir(340);
pair MM = dir(270);
pair D = IP(A--MM, B--C);
pair P = IP(D--D+B-MM, A--B);
pair Q = IP(D--D+C-MM, A--C);
path g = circumcircle(A, P, Q);
pair M = B * 2 - D;
pair N = C * 2 - D;
pair I = IP(L(M, P, 10), g);
pair J = IP(L(N, Q, 10), g);
pair K = IP(M--I, N--J);
path h1 = circumcircle(A, M, C);
path h2 = circumcircle(A, N, B);
pair X = IP(L(A, K, 100), circumcircle(I, J, K));
dot("$A$", A, dir(90));
dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
dot("$D$", D, dir(240));
dot("$P$", P, dir(160));
dot("$Q$", Q, dir(20));
dot("$M$", M, dir(200));
dot("$N$", N, dir(-20));
dot("$I$", I, dir(90));
dot("$J$", J, dir(105));
dot("$K$", K, dir(75));
dot("$X$", X, dir(105));
draw(g, blue);
draw(B--A--C);
draw(M--N);
draw(h1 ^^ h2, heavygreen + dashed);
draw(M--I ^^ N--J, magenta);
draw(circumcircle(K, I, J), heavyred);
draw(D--X, lightolive + dashed);
draw(circumcircle(X, Q, D), orange + dashed);
draw(circumcircle(X, P, D), orange + dashed);
[/asy]](http://latex.artofproblemsolving.com/6/e/3/6e3c95c97b8afe62f4c6dc8cc38236e2801f43b1.png)
and 
are quadruplets of concyclic points. After that, it is a simple angle chasing.
be lines. For each 
,
be distinct points on 
,
.
are concyclic. Then 
are concyclic if and only if 
.
.
if and only if 
.
are tangent at 
centered at 
and 
.
.
in 
then
are collinear.
Hence 
and 
.
,
and 
are cyclic. Hence 
so 
are collinear. 
Hence 
are concyclic. By Reim's theorem 
.
we have 
.
This implies that quadrilateral 
is harmonic, ao 
.
as desired.
,
and 
intersecting ![\[ \measuredangle TMD = \measuredangle TMN = \measuredangle TPQ = \measuredangle TSQ = \measuredangle TSK = \measuredangle TXK = \measuredangle TXD \]](http://latex.artofproblemsolving.com/1/0/8/1086c87fdad340760891ec378adb860d4bc47f33.png)
so 
is cyclic, whence 
,
.
.![\[ \frac{QA}{AX} = \frac{QA}{AD} = \frac{DQ}{BD}, \]](http://latex.artofproblemsolving.com/c/3/1/c3199aa3ef7889d9f20bc050d39d01cdeeb89502.png)
and ![\[ \measuredangle XAQ = \measuredangle DAQ = \measuredangle CDQ = \measuredangle BDQ, \]](http://latex.artofproblemsolving.com/a/8/6/a86485632b98b61176d4c30415cdf465e90bf671.png)
so we get 
.
is cyclic, and now we note that ![\[ \measuredangle XBQ = \measuredangle XDQ = \measuredangle ADQ = \measuredangle APQ = \measuredangle ABC = \measuredangle PBC, \]](http://latex.artofproblemsolving.com/0/f/9/0f9900495c62ac9fdaa37cfe8d1fd4743b595322.png)
so 
and 
are isogonal in 
,
and 
are isogonal in 
:
is the angle bisector of 
.
and 
is tangent to 
hence 
is the midpoint of the arc 
not containing 
is the angle bisector
:
be the circumcenter of 
.
and 
which gives the result
:![\[1=\frac{BD}{DC}.\frac{AC}{AB}=\frac{BM}{BP}.\frac{CQ}{CN}=\frac{\sin BPM}{\sin PMB}.\frac{\sin CNQ}{\sin NQC}=\frac{\sin APK}{\sin KPQ}.\frac{\sin PQK}{\sin KQA}=\frac{\sin KAP}{\sin QAK}\]](http://latex.artofproblemsolving.com/e/2/7/e27975cf3fc09fc284649201609238b19e067a5d.png)
Thus, 
is the angle bisector of 
are collinear
:
![\[\angle IMD=\angle IPQ=\angle IJQ=\angle IJK=\angle IXK=\angle IXD\]](http://latex.artofproblemsolving.com/3/e/a/3ea17d8c32f1e3a2a4ad8c2de04cb3a3178c47e2.png)
![\[\angle DNJ=\angle PQJ=\angle PIJ=\angle KIJ=\angle KXJ=\angle DXJ\]](http://latex.artofproblemsolving.com/f/6/f/f6fb7d8182f8eb8d227bbc6bc72aa544af04c53b.png)
These give that 
are cyclic
:
and 
.
and 
By dividing these we get 
Similarily we have 
and 
By dividing these we get 
.
:
.
.
and 
.
which proves the result since 
![\[\frac{TA}{AD}=\frac{TA}{AX}\overset{?}{=}\frac{TK}{KC}\implies \frac{DK}{AP}=\frac{TK}{TA}\overset{?}{=} \frac{KC}{AD}\]](http://latex.artofproblemsolving.com/b/c/4/bc4fb4073daab0d02393ad4c895235fa10e3cd1e.png)
and 
hence![\[\frac{DP}{DC}=\frac{DK}{KC}\overset{?}{=}\frac{AP}{AD}\]](http://latex.artofproblemsolving.com/a/a/5/aa565d341a1dba7c52930d4e549c8dc508c09144.png)
which is true. Thus, 
are cyclic and similarily we get that 
are cyclic. 
:
.![\[\angle PXB=\angle DXB-\angle DXP=\angle DQB-\angle DCP\]](http://latex.artofproblemsolving.com/2/2/8/2281615bb0ad16f3bb9bdaccacc91ba20e76a296.png)
![\[\angle CXQ=\angle CXD-\angle QXD=\angle CPD-\angle QBD\]](http://latex.artofproblemsolving.com/9/1/3/9131201e98e51beb12f80e8e101f9cfda67f0a85.png)
As desired
,
,
,
,
be the midpoints of 
,
,
,
,
,
.
lies on 
.
and 
.
and similarly, 
.
and point 
are isogonal conjugate, implying that 
.
,
is cyclic. Thus, 
,
.
.
and 
,
,
.
Subtracting these two equations gives 
.
be the midpoint of minor arc 
by shooting lemma. Note that if 
and so 
means 
,
intersect at a dilation centre ![\[\frac{AK}{KT} = \frac{AP}{PT} \frac{\sin \angle KPA}{\sin \angle TPK} = \frac{AB}{BD} \frac{\sin \angle MPB}{\sin \angle BMP} - \frac{AB}{BP},\]](http://latex.artofproblemsolving.com/8/9/7/8973977fb7b42c8ed3e1dbbe87e0324e866a29a4.png)
but 
and 
so 
.
so 
;
by dilation at 
where 
.
,
and this rearranges to 
and so we get 
by dilation at 
we get 
which is the key idea. This gives us Miquel diagram so that 
is cyclic and now it's an easy angle chase.
,
.![\[\frac{AK_1}{K_1D}=2\frac{BP}{PA}=2\frac{CQ}{QA}=\frac{AK_2}{K_2D}\]](http://latex.artofproblemsolving.com/b/0/6/b06fdc0cecdb0cbd89c6e00450d8e0cf0a37da82.png)
by addendo on parallel lines.
,
cyclic
.
to points such 
,
are isosceles trapeziums. In fact, as 
,
is parallel to 
hence we actually have the two trapeziums congruent.
is the midpoint of 
,
midpoint of 
,
,
.
as midpoints of 
,
from parallels and trapezium, hence they work.
over 
,
collinear. By Reim from parallel lines & homothety 
is cyclic. Dilating factor half at 
as the midpoint of 
,
cyclic. Now the collinearities are trivial, say by 
rotation at 
isogonal, or 
,
.
bisects 
.
is sent to 
via a homothety centered at 
via a homothety centered at 
.
at 
is cyclic, by Reim's theorem again, 
is cyclic; since both quadrilaterals 
and 
,
-
in 
.
,
is mapped to the reflection of 
is mapped to the reflection of 
maps to line 
under our inversion.
is mapped to the reflection of 
is the reflection of 
passing through 
,
passing through 
.
and 
are the second intersections of 
respectively, while 
and 
are the second intersections of 
as well as 
,
.
so 
.
is the angle bisector of 
.
so 
.
lies on 
.
so 
is tangent to 
is tangent to 
so 
and 
Note that 
.
and 
.
so 
.
and 
at 
.
and 
are cyclic.
.
and 
are cyclic.
and 
so 
is parallelogram and since 
is also parallelogram so 
so 
is cyclic. we can similarly prove that 
and 
are parallelogram and so 
are cyclic, we have that 
is cyclic. Now note that 
.
be the circumcircle of 
be the intersections of 
with 
respectivly and let 
be the intections of 
with 
respectivly.
.
to 
.
.
.
.
and 
and 
we get that 
and 
we get that 
is parallel to ![\[
\frac{KL\cdot KI}{KR\cdot KJ}=\frac{KP\cdot KI}{KQ\cdot KJ}=1
\]](http://latex.artofproblemsolving.com/f/5/8/f58bc71718892f14c746127c6835884d0874dfca.png)
and 
is a cyclic quadrilateral. So we have![\[
\frac{DE}{DL}:\frac{DF}{DR}=\frac{DE\cdot DJ}{DL\cdot DJ}:\frac{DF\cdot DI}{DR\cdot DI}=\frac{DE\cdot DJ}{DF\cdot DI}:\frac{DL\cdot DJ}{DR\cdot DI}=1
\]](http://latex.artofproblemsolving.com/b/5/f/b5fded65b8ba65e77c0eae50f7cbbc9364dc9ec3.png)
and thus 
.
we get that the tangent at 
is tangent to 
and 
we get that 
and 
.
is cyclic and since 
we have 
.
.
.
.
with 
.
is the midpoint of 
and the circumcircles of 
and 
are externally tangent at 
and thus 
.
has the same inscribed angle w.r.t. the circumcircles of 
and 
.
and 
.
and 
are two circle with identical radii so that the arcs 
and 
have the same length since they are symmetric w.r.t. 
.
is cyclic.
.
whence the claim follows. 
are cyclic.
which suffices. 
proving the claim. 
.
as,
and we have that,
as desired. In an identical manner we may prove 
so that,
as desired. Thus 
so the result follows. 
It does feel like two different problems combined into one though, because once you see that 
,
.
followed by a reflection around 
to 
,
,
by angle chasing, so 
.
,
.![\[ \angle BXP = \angle XPA - \angle XBA = \angle CXA - \angle QXA = \angle CXQ. \]](http://latex.artofproblemsolving.com/f/2/3/f23f4ffdf4ee18629c9c75e93aa2ed3547e69cee.png)
Next, we claim that 
By trig Ceva in 
.
we have to show that 
which is true since 
,
is cyclic. Thus, by Reim's again, 
,
,
,
is cyclic, or 
.
,
.
,
,
and 
,
and 
intersect at 
would be a rectangle and 
and 
is 
and the power of 
is 
,
is cyclic as 
Thus 
,
be an inversion about 
and 
.
,
is mapped to 
,
in a plane. denote the vectors 
by 
respectively. make 
.
So necessarily and sufficiently we need to prove
Construct the projections of X w.r.t lines AB and AC and denote them with U, V, respectively. Then XU=XV because AX bisects 
.
Thus, only the statement below need to be proved:
Similar to other solutions, it's proved that 
.
Thus
The similar conclusion also holds for AC. So
That's what we desired.
,
,
,
is cyclic. Since 
,
is cyclic, and similarly 
is cyclic. Let 
be the radical axis of 
,
,
from 
.
be the second intersection of 
.
,
,
is tangent to 
.
,![\[\angle GNJ = \angle GAJ = \angle BAD + \angle BAJ = \angle CAD + \angle CAN = \angle GAN = \angle GJN\]](http://latex.artofproblemsolving.com/7/f/3/7f3bfb4a31f3e3a10749a62ea3b3dec2def5bc43.png)
. This gives 
,
.
,
.![\[\angle BJQ = \angle BJN = \angle BAN = \angle BAC + \angle CAN = \angle BAC + \angle BAJ = \angle JAC = \angle JAQ,\]](http://latex.artofproblemsolving.com/9/2/5/9258ee7f94e4d1f38c50cc8eb9f2abf037ac6c12.png)
,
,
,
is cyclic, and similarly 
is cyclic. Let 
be the second intersection of 
with 
be the second intersection of 
with ![\[ \angle APK = \angle API = 180^\circ - \angle AIC = \angle AEC = \angle AEK, \]](http://latex.artofproblemsolving.com/e/a/c/eac0599e8174d55e6c00a5b0277cd13cc4c9426b.png)
so 
,
,
,
is cyclic, and since ![\[\angle BFQ = \angle KFQ = \angle KAQ = \angle PAD = \angle PQD = \angle CDQ = 180^\circ - \angle BDQ,\]](http://latex.artofproblemsolving.com/f/8/5/f85cd384ad4d03535244cfb59de2b444ada0bedf.png)
is cyclic, and similarly 
is.
,
is tangent to 
(containing 
,
is cyclic, and since 
,
.
and similarly 
.
,
and 
and 
.
,
are concyclic.
and 
,
.
,
is the midpoint of 
to denote the inverted image of any point 
,
.
lies on 
and 
.
and 
,
is a parallelogram. Thus 
are conyclic.
,
.
,
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/1/c/4/1c4a176023291ebb444f8f36e3b5308417be9501.png)
are collinear.
collinear.
which means that 
and 
are homothetic, so 
collinear. 
.
,
so 
is cyclic, which implies
so 
,
and let 
meet 
at 
.
.
so 
.
is a parallelogram, and so 
are collinear.
![[asy]
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draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle((12.60864,4.52943), 15.73881), linewidth(0.6)); draw(A--D, linewidth(0.6)); draw(B--M, linewidth(0.6)); draw(C--N, linewidth(0.6)); draw(M--I, linewidth(0.6)); draw(N--J, linewidth(0.6)); draw(circle((9.84830,26.28453), 24.79711), linewidth(0.6)); draw(X--A, linewidth(0.6)); draw(circle((-3.50455,16.62872), 32.09345), linewidth(0.6) + linetype("4 4") + blue); draw(circle((23.45318,22.66332), 35.60940), linewidth(0.6) + linetype("4 4") + red);
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is done by taking a homothety that maps the midpoint of arc 
of 
not containing 
and 
.
and 
.
with ![\[ \frac{AP}{PB}\cdot \frac{BM}{MD}\cdot \frac{DK'}{K'A}=-1 \implies \frac{DK'}{K'A} = - \frac{PB}{AP}\cdot (-2) = 2\cdot \frac{PB}{AP} .\]](http://latex.artofproblemsolving.com/a/f/6/af60f912d35f8641a372099dc1c033e64f1ce1ef.png)
![\[ \frac{DK''}{K''A} = 2\cdot \frac{QC}{AQ} .\]](http://latex.artofproblemsolving.com/4/d/3/4d3d081980c2bb24ed5d8b1e94ec1446f914cc28.png)
.
,
.
.
![\[ \measuredangle DXI=\measuredangle KXI=\measuredangle KJI =\measuredangle NJI=\measuredangle NMI=\measuredangle DMI .\blacksquare\]](http://latex.artofproblemsolving.com/0/1/f/01fd69a3be6e8b5b605b499d44306dcc722a2bac.png)
.![\[ \measuredangle MXD=\measuredangle MID=\measuredangle PID =\measuredangle PAD=\measuredangle BAD \]](http://latex.artofproblemsolving.com/5/3/b/53b85ad938f47718261a0c566757355b0ea4b556.png)
which gives us that 
.
,
.![\[ \measuredangle CDP=\measuredangle BDP=\measuredangle DQP =\measuredangle DAP=\measuredangle QAD=\measuredangle QAX .\]](http://latex.artofproblemsolving.com/9/f/8/9f803eb1bcf0782c30581b09bf5c768d820f372a.png)
.
.
due to the tangency. This gives us that ![\[ \frac{CD}{CA}=\frac{DP}{DA}=\frac{DP}{AX}\implies \frac{CA}{AX}=\frac{CD}{DP} .\]](http://latex.artofproblemsolving.com/2/0/6/206594b2ce951d76f7c8eb9eaed05b3f0eade8d2.png)
.![\[ \measuredangle DXC=\measuredangle AXC=\measuredangle DPC \]](http://latex.artofproblemsolving.com/4/8/7/4879ea9a0281a477eb8701d9f2a46c6d05a1ad57.png)
which gives us our desired claim. ![\[ \measuredangle BXP=\measuredangle BXD-\measuredangle PXD = \measuredangle BQD-\measuredangle PCD =\measuredangle BQD-\measuredangle CPQ .\]](http://latex.artofproblemsolving.com/1/7/0/1705172a7825c3411fda1ebe2f6ca96ce657075e.png)
.
which is indeed true and we are done.
![\[\frac{\sin(\angle AQJ)}{\sin(\angle API)}\cdot\frac{\sin(\angle QPI)}{\sin(\angle API)}=1\]](http://latex.artofproblemsolving.com/4/3/4/4347bd679473a41add97345a21bc347f12b0001b.png)
But this is equal to ![\[\frac{\sin(\angle CQN)}{\sin(\angle CNQ)}\cdot\frac{\sin(\angle PMB)}{\sin(\angle MPB)}=\frac{CN}{CQ}\cdot\frac{BP}{BM}=\frac{DC}{CQ}\cdot\frac{BP}{BD}=\frac{AC}{CD}\cdot\frac{AB}{BD}=1\]](http://latex.artofproblemsolving.com/4/7/1/47102e937df151994480ca1f044916ec78f36468.png)
meet ![\[-1=(M,D;B,\infty_{BC})\overset{P}{=}(I,D;A,Q)\]](http://latex.artofproblemsolving.com/d/f/1/df1d450b4ef7cdbc9e4684d266c00a6c7238cb6a.png)
So 
or 
.![\[TI\cdot TN=TD^2=TJ\cdot TM\]](http://latex.artofproblemsolving.com/2/f/c/2fc19ede601dc769aa2b086362731e752934848d.png)
So ![\[\frac{SJ}{SI}=\frac{AJ}{AI}\cdot\frac{TJ}{TI}\]](http://latex.artofproblemsolving.com/b/3/d/b3d026bd829a984d35d9ca60673c7d111851f6cb.png)
Now we have 
and ![\[\frac{AJ}{AI}=\frac{\sin(\angle ADJ)}{\sin(\angle ADI)}=\frac{\sin(\angle DJN)}{\sin(\angle NJI)}\cdot\frac{\sin(\angle JIM)}{\sin(\angle MID)}=\frac{\sin(\angle KNM)}{\sin(\angle KMN)}=\frac{KM}{KN}\]](http://latex.artofproblemsolving.com/5/3/7/537eb040ac98bae21d7beebae73a3ce034d2474b.png)
Now it remains to prove that 
which is true since 
.
.![\[\frac{JD}{ID}=\frac{XJ}{XI}=\frac{\sin(\angle XKJ)}{\sin(\angle XKI)}=\frac{\sin(\angle DKN)}{\sin(\angle DKM)}=\frac{DN}{DM}\cdot\frac{KM}{KN}=\frac{DN}{DM}\cdot\frac{JM}{IN}\]](http://latex.artofproblemsolving.com/5/f/c/5fcef1de4ec79f60137058898fa295f7e409a3aa.png)
This is true since 
and 
.
is tangent to 
so ![\[SX^2=SJ\cdot SI=SD^2\]](http://latex.artofproblemsolving.com/c/8/9/c89c64983685094084fbf6baeb05a4eacb6c8bd6.png)
And since 
we have that 
and ![\[\frac{AX}{AB}=\frac{AD}{AB}=\frac{PD}{BD}=\frac{DQ}{BD}\]](http://latex.artofproblemsolving.com/b/c/4/bc427e9734ba81dea4ef3ac0a58e6396d3b8bf94.png)
Therefore 
so 
and similarly 
so the conclusion follows. ![[asy]
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lies on the circle with diameter 
,
.
.
with transversal 
,
,
.
,
.
since that 
,
,
.
,
and 
,
and 
are parallelograms. Thus 
.
and 
are cyclic. After that, we can derive 
,
.
.
,
.
and 
are isogonal w.r.t. 
,
collinear.
and 
the intersection of 
and 
.
gives 
so 
.
Which gives 
cyclic. But 
is cyclic and we are done by Reim.
.
And 
can be easily obtained from: 
.
and substracting 
gives the desired result.
be the arc midpoint of 
not containing 
are collinear.
![\[\frac{\sin\angle APJ}{\sin\angle \angle JPQ} \cdot \frac{\sin\angle IQP}{\sin\angle PQI} = \frac{\sin\angle MPB}{\angle PMB} \cdot\frac{\sin \angle QNC}{\angle CQN}= \frac{PB}{MB}\cdot\frac{CN}{CQ} = \frac{PB}{BD}\cdot\frac{CD}{CQ}\]](http://latex.artofproblemsolving.com/b/6/d/b6da56393a5501eeca1f0ec1e59a03f860a3a9f9.png)
However, 
so this evaluates to 1, and so 
bisects 
so 
which implies 
Finally, observe that a 
homothety centered at 
Let 
tangent to 
at 
Show that 
are equal.
Hence, 
We also have 
while
so
while![\[\overline{\frac{(ap-1)(a-p)}{(a-1)^2p}} = \frac{(ap-1)(a-p)}{(1-a)^2p} \implies \frac{(ap-1)(a-p)}{(1-a)^2p}\in\mathbb R\]](http://latex.artofproblemsolving.com/9/2/8/928e3f60d93b89203188bf549b5dba410138b4cc.png)
which finishes
are tangent to 
but 
)
Let 
.
and 
and 
.
and 
be midpoints of 
and 
.
and hence if we 
invert at 
and 
which is obviously parallel to 
.
homothety at 
concur. See that 
and hence 
and similarly 
.![\[-1=(P,D;P',\infty) \overset B= (A,D';\overline{BP'} \cap \overline{AD},M_A)\]](http://latex.artofproblemsolving.com/0/c/5/0c57e85c3f2d1316d98f9fecc69c7ed51a504c4f.png)
And similarly 
and done. 
.![\[(X,D;A,\infty)=(F,A;D,K) \overset P= (B,\infty;M,D)=-1\]](http://latex.artofproblemsolving.com/6/a/c/6ac4b9e26fa0f85528b924829d434fef02ba6308.png)
And done. 
is cyclic (and so is 
)
be midpoint of 
.
and hence see that 
and hence ![\[\angle AQ'D=180^{\circ}-\angle AQ'Q=180 ^{\circ}-\angle AF'F=180 ^{\circ}-\angle AB'D\]](http://latex.artofproblemsolving.com/5/5/8/5586c6a4f31f5918e1593ebf35a971f3616c48b3.png)
And done. 
and hence 
and 
are isogonal in 
and 
are isogonal in 
![\[\angle DPQ = \angle PDB = \angle DQP\]](http://latex.artofproblemsolving.com/8/a/5/8a547ba5447dd12efc1de747b9109bc236ea8cff.png)
Thus 
which tell us 
![\[ \angle IMB = \angle IPQ = \angle IJQ = \angle IJN\]](http://latex.artofproblemsolving.com/f/e/3/fe356c7cd8d930cc870d983d0a9f9f1e7fea0958.png)
thus 
cyclic.
and 
.
tangent to ![\[\angle AJI = \angle API = \angle MPB = \angle AMB\]](http://latex.artofproblemsolving.com/e/0/0/e00ce171935c3e12ebcb529411d156f3b679ccb2.png)
and similarly 
,
,
.
and 
is 
only.
inversion with reflection about angle bisector 
, 
,
.
.
we get 
and 
(As 
lie on line 
and 
.
![\[ \angle KNM = \angle JIK = \angle JXA = \angle ANX' \]](http://latex.artofproblemsolving.com/0/c/e/0ce3ee281c64127154f55317323cb4b7c5951d05.png)
and similarly 
.
.
are collinear, we get 
which implies 
.
.
.![\[\angle AIX + \angle AXI = \angle AIX + \angle JIP = \angle IAL = \angle JIL = \angle PIL + \angle JIP \]](http://latex.artofproblemsolving.com/6/d/5/6d5d51468643f18ff3159b9cb05124a50f1636da.png)
which implies 
and similarly 
but as 
.
we get 
,
.
.
.
, 
,
where 
is tangent to 
and 
respectively. Notice if 
.![\[\angle BX'P = \angle BX'A - \angle PX'A = \angle DCH - \angle DP'H\]](http://latex.artofproblemsolving.com/1/4/f/14f62f05b63f9d3510a1b3c7098f0e8802bcd979.png)
and 
.
and 
,
with 
and 
.
.
and 
.
,![\[\angle BND = \angle DUN + \angle NCD = \angle ABD - (\angle NBD - \angle NCD)\]](http://latex.artofproblemsolving.com/6/b/d/6bddc88da58ee1a7089fdd1788292b38b000ae04.png)
But we also have 
give implies ![\[ \angle BND = \angle NVA + \angle ABN = \angle ACD - (\angle ACN - \angle ABN)\]](http://latex.artofproblemsolving.com/0/f/a/0fa59e64071189988677167f3b199f33b7b6ec66.png)
As 
cyclic we get 
,
which is equivalent to 
,
