let ABC be a triangle such that AB<BC , AC not equal to BC and K be its circumcircle. The tangent line to K at the point A intersects the line BC in the point D. Let k be the circle tangent to K and to the segment AD and BD. We denote by M,N the points where k touches BD and AD respectively. Let J be the center of the exscribed circle which is tangent to the side AB. Prove that J,M,N are collinear.

We need Darij to check it

It seems that Darij has a competitor! As far as I remember, Darij's longest post was 7 screens (!) long (on 1024 x 768), and this one is 6 but without LaTeX formatting...

Did you expand all hide-scripts ?

From 's theorem applied to points B,A, circle k, and point C we have the relations:
.(1)
Let intersect at . Then from Menelaus’s theorem applied to triangle BAD we have:
,
but hence,
and comparing it with (1) we obtain:
and
from the converse of 's theorem we get that is on , hence on .

Hi, it's Darij speaking. ML is failing to load on my computer right now, so I have asked a friend of mine to submit this message (thanks!).
____________________________________________________________________

This thread is old, but maybe someone is still interested in a synthetic and elementary solution I found three days ago on the Rostock seminar:

Preliminaries: 1. We will use directed angles modulo 180° throughout the following text.
2. The abbreviation "circle " always means the circle through three given points , and .

I reduced the problem to the following theorem:

Theorem 1. Let ABC be a triangle. Let D, E, F be the midpoints of its sides BC, CA, AB, and let A', B', C' be the feet of the altitudes to these sides. Also, let H be the orthocenter of triangle ABC, and let D', E', F' be the midpoints of the segments AH, BH, CH. It is pretty well-known that the points D, E, F, A', B', C', D', E', F' lie on the nine-point circle of triangle ABC.

Now, let the line AE' intersect this nine-point circle at a point E" (apart from the point E'). Let the tangent to the nine-point circle of triangle ABC at the point B' meet the line C'A' at a point S. Finally, let the perpendicular to the line CA at the point A meet the line B'S at a point V and the line C'A' at a point W.

Then,

(a) the circle VWE" touches the line B'S at the point V, the line C'A' at the point W, and the nine-point circle of triangle ABC at the point E";

(b) the points W and E both lie on the perpendicular to the line AE' at the point E".

Proof. Since < CC'A = 90° and < CA'A = 90°, the points C' and A' lie on the circle with diameter CA. Thus, < AC'A' = < ACA'. On the other hand, since the point H is the orthocenter of triangle ABC, the lines AH and BH are altitudes of triangle ABC, and thus they are perpendicular to its sides BC and CA, respectively; thus, < (AH; BC) = 90° and < (CA; BH) = 90°. Finally, since the points E' and F are the midpoints of the sides BH and AB of triangle ABH, the line E'F is a midparallel in this triangle, and thus it is parallel to the side AH, so that AH || E'F, and thus < (BH; AH) = < (BH; E'F). Combining these results, we find that

< AC'W = < AC'A' = < ACA' = < (CA; BC) = < (CA; BH) + < (BH; AH) + < (AH; BC)
= 90° + < (BH; AH) + 90° = 180° + < (BH; AH) = < (BH; AH)
= < (BH; E'F) = < B'E'F.

Since < AB'B = 90°, the point B' lies on the circle with diameter AB. Of course, the points A and B also lie on this circle with diameter AB. The center of this circle is clearly the midpoint F of the segment AB. Thus, B'F = AF. Hence, the triangle AFB' is isosceles, and thus, < B'AF = < FB'A. In other words, < (CA; AB) = < (FB'; CA). Now, we know that the line BB' is perpendicular to the line CA (as an altitude in triangle ABC); thus, < (CA; BB') = 90°. On the other hand, since the point W lies on the perpendicular to the line CA at the point A, we have , so that < (AW; CA) = 90°. Thus,

< WAC' = < (AW; AB) = < (AW; CA) + < (CA; AB) = 90° + < (CA; AB)
= 90° + < (FB'; CA) = < (CA; BB') + < (FB'; CA) = < (FB'; BB') = < FB'E'.

Thus, we have found that < AC'W = < B'E'F and that < WAC' = < FB'E'. Consequently, the triangles AC'W and B'E'F are directly similar. Thus, . In other words, .

We know that the points A, V, W lie on one line - namely, on the perpendicular to the line CA at the point A. Let W' be the orthogonal projection of the point E' on this line. Then, < E'W'A = 90°. On the other hand, < W'AB' = 90°, since the point W' lies on the perpendicular to the line CA at the point A. Finally, < AB'E' = 90°, since the point E' lies on the altitude BB' of triangle ABC. Hence, the quadrilateral W'AB'E' has three right angles; therefore, it is a rectangle. Consequently, B'E' = AW'. We also know that B'F = AF. Thus, the equation we obtained before becomes . But since the points F, C', E' and E" lie on one circle (namely, on the nine-point circle of triangle ABC), the intersecting chords theorem yields . Thus, we get . By the converse of the intersecting chords theorem, this implies that the points W, W', E' and E" lie on one circle. Since the point W' is the orthogonal projection of the point E' on the line through the points A, V, W, we have < E'W'W = 90°, and thus, the segment E'W is a diameter of the circle through the points W, W', E' and E". Consequently, < E'E"W = 90°. Thus, the point W lies on the perpendicular to the line AE' at the point E".

On the other hand, since the line BB' is an altitude in triangle ABC, we have < (CA; BB') = 90°, or, in other words, < EB'E' = 90°. Remembering that the points E, B' and E' lie on the nine-point circle of triangle ABC, it becomes clear that the equation < EB'E' = 90° yields that the segment EE' is a diameter of the nine-point circle. Hence, < EE"E' = 90° (as the point E" also lies on the nine-point circle). In other words, the point E lies on the perpendicular to the line AE' at the point E".

Thus, altogether, we have shown that the points W and E both lie on the perpendicular to the line AE' at the point E". This proves Theorem 1 (b). Now, we will show Theorem 1 (a):

Let the circle VWE" meet the line AE' at a point R (different from the point E"). Then, since the point W lies on the perpendicular to the line AE' at the point E", we have < WE"R = 90°, and this shows that the segment WR is a diameter of the circle VWE". In other words, the circle VWE" is the circle with diameter WR. On the other hand, since the segment EE' is a diameter of the nine-point circle of triangle ABC, the nine-point circle of triangle ABC can be seen as the circle with diameter EE'.

Since the line B'S is the tangent to the nine-point circle of triangle ABC at the point B', by the tangent-chordal angle theorem, we have < (B'S; B'E") = < B'E'E". On the other hand, the lines BB' and AV, both being perpendicular to the line CA, are parallel to each other, so that < (BB'; AE') = < (AV; AE'). Thus,

< VB'E" = < (B'S; B'E") = < B'E'E" = < (BB'; AE') = < (AV; AE') = < VAE".

Thus, the points V, E", B' and A lie on one circle. Hence, < E"VA = < E"B'A. In other words, < E"VW = < E"B'E. But since the point R lies on the circle VWE", we have < E"VW = < E"RW, and since the points E", E, B' and E' lie on one circle (namely, on the nine-point circle of triangle ABC), we have < E"B'E = < E"E'E. Thus, the equation < E"VW = < E"B'E becomes < E"RW = < E"E'E. In other words, < (AE'; WR) = < (AE'; EE'). Hence, WR || EE'.

Now, let's look once more at our configuration: The points W, E and E" lie on one line, the points R, E' and E" lie on one line, and we have WR || EE'. Thus, there exists a homothety with center E" which maps the segment WR to the segment EE'. This homothety then, of course, must map the midpoint of the segment WR to the midpoint of the segment EE'. Since the point E" is the center of this homothety, and since the center of a homothety always lies on the line joining a point with its image under that homothety, it thus follows that the point E" lies on the line joining the midpoints of the segments WR and EE'. Since the midpoint of a segment is the center of the circle which has this segment as diameter, we can restate this result as follows: The point E" lies on the line joining the centers of the circles with diameters WR and EE'.

Remembering that the circle with diameter WR is the circle VWE" and that the circle with diameter EE' is the nine-point circle of triangle ABC, we thus can state: The point E" lies on the line joining the centers of the circle VWE" and of the nine-point circle of triangle ABC. But on the other hand, we know that the point E" lies on both of these circles, i. e. the point E" is a common point of these two circles. It is well-known that, if a common point of two circles lies on the line joining the centers of these circles, then these two circles touch each other at that point. Hence, we conclude that the circle VWE" and the nine-point circle of triangle ABC touch each other at the point E". In other words, we have shown that the circle VWE" touches the nine-point circle of triangle ABC at the point E".

In order to complete the proof of Theorem 1 (a), it remains to show that the circle VWE" touches the line B'S at the point V, and the line C'A' at the point W.

It is well-known that the perpendicular bisector of a chord in a circle always passes through the center of the circle. Thus, since the points C' and A' lie on the circle with diameter CA, the perpendicular bisector of the segment C'A' passes through the center of the circle with diameter CA, i. e. through the midpoint E of the segment CA. Similarly, since the points C' and A' lie on the circle with diameter BH (this is because < BC'H = 90° and < BA'H = 90°), the perpendicular bisector of the segment C'A' passes through the midpoint E' of the segment BH. Thus, the perpendicular bisector of the segment C'A' passes through both points E and E'. In other words, the perpendicular bisector of the segment C'A' is the line EE'. Hence, the line EE' is perpendicular to the line C'A'. Since WR || EE', the segment WR must also be perpendicular to the line C'A'. In other words, the line C'A' is the perpendicular to the segment WR at its endpoint W. But the segment WR is a diameter of the circle VWE", and the perpendicular to a diameter of a circle at one of its endpoints is the tangent to the circle at this endpoint; thus, the line C'A' is the tangent to the circle VWE" at the point W. In other words, the circle VWE" touches the line C'A' at the point W.

Remains to show now that the circle VWE" touches the line B'S at the point V.

Since the points V, E", B' and A lie on one circle, we have < E"VB' = < E"AB'. Since the point E lies on the perpendicular to the line AE' at the point E", we have , so that < (AE'; EE'') = 90°, and since the point W lies on the perpendicular to the line CA at the point A, we have , so that < (AW; CA) = 90°. Thus,

< (E"V; B'S) = < E"VB' = < E"AB' = < (AE'; CA)
= < (AE'; EE") + < (EE"; AW) + < (AW; CA) = 90° + < (EE"; AW) + 90°
= 180° + < (EE"; AW) = < (EE"; AW) = < E"WV.

Thus, by the converse of the tangent-chordal angle theorem, the line B'S is the tangent to the circle VWE" at the point V. In other words, the circle VWE" touches the line B'S at the point V. This completes the proof of Theorem 1.

Now, let's come to the solution of the problem; we will use the notations from Yetti's post #2. So, we have a triangle ABC and its excenters , , (opposite to the vertices A, B, C, respectively). We will apply Theorem 1 to the triangle ; but before we do this, we recall some simple facts about this triangle:

- The points B, A, C are the feet of the altitudes of the triangle . (This is well-known - and pretty clear: Since the internal and the external angle bisector of an angle are always perpendicular to each other, we have , and , so that the points B, A, C are the feet of the altitudes of the triangle .)
- The circumcircle of triangle ABC is the nine-point circle of the triangle . (This is clear, since it passes through the feet B, A, C of the altitudes of triangle .)
- The midpoints , , of the sides , , of the triangle are the midpoints of the arcs ABC, CAB, BCA on the circumcircle of triangle ABC. (Prove this for yourself - it's pretty easy.)
- The orthocenter of triangle is the incenter I of triangle ABC, and the midpoints , , of the segments , , are the midpoints of the arcs CA, BC, AB on the circumcircle of triangle ABC. (Dito.)

Using these facts, after applying Theorem 1 to the triangle , we get the following results:

Application of Theorem 1 to the triangle :

Let the line intersect the circumcircle of triangle ABC at a point (apart from the point ). Let the tangent to the circumcircle of triangle ABC at the point A meet the line BC at a point D. Finally, let the perpendicular to the line at the point meet the line AD at a point and the line BC at a point .

Then,

(a) the circle touches the line AD at the point , the line BC at the point , and the circumcircle of triangle ABC at the point ;

(b) the points and both lie on the perpendicular to the line at the point .

Well, what follows from these results? Very much. In fact, the result (a) shows that the circle is tangent to the circumcircle of triangle ABC and to the lines AD and BD. But there exist exactly three circles tangent to the circumcircle of triangle ABC and to the lines AD and BD - namely, the circles , and . Hence, the circle must coincide with one of these three circles , and . Actually, it is clear that the circle cannot coincide with the circle (since the circle touches the line AD at the point A, while the circle touches the line AD at the point which is clearly different from A). Hence, the circle coincides with one of the two circles and . Looking at the arrangement of points, one concludes that the circle coincides with the circle . Hence,

- the point , at which the circle touches the line AD, coincides with the point , at which the circle touches the line AD;
- the point , at which the circle touches the line BD, coincides with the point , at which the circle touches the line BD;
- the point , at which the circle touches the circumcircle of triangle ABC, coincides with the point , at which the circle touches the circumcircle of triangle ABC.

From here we can draw a lot of conclusions:

- We know that the points and lie on the perpendicular to the line at the point . Substituting and , we can conclude that the points and lie on the perpendicular to the line at the point . In particular, this yields that the points and lie on one line with the point . This solves the initial problem.
- The point was initially defined as the point where the line intersects the circumcircle of triangle ABC. Hence, the point lies on the line . Keeping in mind that , we conclude that the point lies on the line . In other words, the line passes through the point . This solves the "Related problem" proposed by Yetti (note that he uses the notation K for the point which I call ).
- Part (b) of our result above states that the points and both lie on the perpendicular to the line at the point . Since and , this can be rewritten as follows: The points and both lie on the perpendicular to the line at the point . This fact doesn't seem to have been mentioned in this thread until now.

So much for the problem. Two remarks:

1. Thanks, Yetti, for your "Related problem" - it did help me in finding the above solution of the initial one. This gives another nice example of how proving a stronger assertion is easier than proving a weak one.
2. No guys, my longest post on ML was much more than 7 screens long (it takes 12 sides on paper) and didn't contain any hide-scripts. Guess which post I mean. (No, not this one.)

darij

So much for hoping to see a shorter solution to the problem...

Hello Pestich,

Though the analogy with the mixtilinear incircle configuration is striking, I fear that it cannot help to find a new solution of this problem. I don't see how extraversion could make the line AD (the tangent to the circumcircle of triangle ABC at the point A) out of a sideline of triangle ABC. Moreover, the incenter of triangle ABC is the midpoint between the points of tangency of a mixtilinear incircle with two sides of triangle ABC, while in our problem, the excenter is only collinear with the points of tangency, but is far from being their midpoint.

Darij

1. WOW !!! I did not know the solution of the related problem, it just showed on the Sketchpad and it was plausible - what is true for an incircle and one excircle should be true for the remaining two excircles.
2. The guys were laughing at me. It was my 1st post at this forum, I did not know any LATEX at the time, the post was horrible (with a lot of hide scripts) and only latexed at a later time.

Wow, Pestich, a very good - and correct - observation! And it can be proven exactly in the same way as Sailor solved the original problem in post #8. But it can also be proven similarly to Lemma 1 in [1], §2 (actually, it's an extraversion of that lemma). So you were right that extraversion could help!

References

[1] Jean-Louis Ayme, Sawayama and Thébault's Theorem, Forum Geometricorum 3 (2003) pp. 225-229.

darij

Dear Mathlinkers,
working about mixtilinear incircles, I found this interesting problem which I have proved last year on my site
see : http://perso.orange.fr/jl.ayme vol. 2 Un remarquable probleme de Vladimir Protassov p.6 "Un alignement".
Sincerely
jean-Louis