Problem regarding circles?

Question

A triangle ABC with incentre O has its incircle touching it at points K, L and M on AB, BC and CA respectively. A line throough B parallel to KL meets MK and LM extended at R and S respectively. Prove that angle ROS is acute.
I have the diagram too. If you need it, give me your email ID and I'll send you an attachment.

Answer 1

I have been thinking a bit in your problem. I still haven´t solved it, I just got up and I only give it some thoughts, but these could help you. I´m sorry if I will invent some terminology, I hope that you will be able to follow my reasoning though. There are probably much better ways to prove this, and I probably will be proving unnecessary things, but I will be thinking out loud. Let´s see if we can get to the demonstration that you need. I hope that you will be able to follow my reasoning.

1) BO is the bisector of the angle KBL, which is the same than the angle ABC
2) KL is parallel to RS, this is a very important datum, I think.
3) KO is perpendicular to AB, and KO is a radius
4) LO is perpendicular to BC, and LO is a radius
5) From 3) and $) you deduce than LOKB is a quadirilateral, with 2 right angles, so, it's incriptible. It seems to me that RS is perpendicular to BO. But I´m not sure how to prove this yet.
6) Since the angles in K and L are right, then BO is a diameter, and BO is perpendicular to KL, which is too an important fact.
7) KLO is an isosceles triangles, it has 2 radius, and BO is the mediatrix of KL
Let's call H the intersepts KL with BO.
8) OKH is equals to OLH. They are right triangles with a common side and 2 equals hypothenuses, so, because of Pythagorean theorem, their thrid side is equals. So, KH is equals to LH. And LK is perpendicular to BO, so BO is the height of the triangle LBK, and it´s it´s mediatrix, so BKH is an isosceles triangle.
9) From 7 and 8 you deduce than the quadrilateral BLOK is a romboid. It's build of 2 isosceles triangles. And it´s a romboid with 2 right angles.
10) There is a axial symmetry and KR is equals to LS which is important too, since the angles R and S are then equals and RKLS is an isosceles trapeze. So, it'´s incriptible too. This is a property of any isosceles trapeze.
11) BO is the symmetry axis from this isosceles trapeze.
12) The angles is R (or S) are suplementary form the angles RKH (or RLH). Remember that the lines KL and RS are paralell ones. I think that we are pretty close now.
13) Let´s work now with the triangles KOR and LOS, which are equals because of the symmetry. They have 2 equals pair or sides (KR=LS and the 2 radius), and they have the by these pair of sides formed angles equals: RKO is equals to SLO because they are the sum of 2 equals pairs of angles: OKL = OLK and RKL = SLK
14) Since these triangles are equals, then the sides SO and RO are equals. So, ROS is an isosceles triangle too. And BO is the height of it since it´s the symmetry axis from the triangle.
15) Since RKLS is an isosceles trapeze, then the point M belongs to the symmetry axis, so M, O and B are in the line BS, which is important too.
16) So, SMR is an isosceles triangle too.
17) Since MO is a radius and AC is tangent to this radius, then AC is perpendicular to BO, which is the same line than BM or than MO. This is important too.
18) Hence, AC is parallel to the side KL, because the two of them are perpendicular to the same straight line. And,since KL is perpendicular to RS, then AC is perpendicular to RS too.
19) Since AC is perpendicular to AM, then AM is the height of the triangle and it´s the mediatrix too due the symmetry of the problem. And, since M is the middle point of AC, then BM is the mediana.
20) MO is a radius, so, it´s known too.
21) KLM is an isosceles triangle too because of the same symmetry. MO is its height and MO is the mediana of this triangle. Remember that MO is the same than MH.
22) Let's now think of the quadrilateral JKML, where J is the intersepts point MO with the circle. It´s inscriptible too, and it´s incripted in the circle that you draw. And, since KL is perpendicular to HM, then the H is the middle point from JO. And JO is equals to MO, since both of them are radius, so HO is the hafl from OM, and then O is the baricenter of the triangle MKL. This is important too.
23) If we can prove than KLM is an equilateral triangle, then the angle M will be 60o and the triangle SMR will be equilateral too (remember that it´s isosceles).
24) O is the baricenter, and its the circumcenter from this triangle. And, since KO, LO and MO are radius, then the 3 medianas measure exactly the same and LKM is an equilateral triangle, so it´s RML

These facts seem important but I still haven´t proved what you wanted.

25) Let´s come back to the triangles KOR and LOS. We already proved that they were equals (13). RBO and SBO are equals too, because of the symmetry. They are right triangles and the angles ORB, OSB are equals and acute, since they are less than BRM and BSM which are 60o because SRM is an equilateral triangle.
26) Since these triangles are right, then the angles BOR and BOS are acute too, and they are equals to each other. The sum of them is the angle that you wanted to prove that it is acute. But I'm stucked here, I´m sorry. I can´t realise how to prove that this sum is an acute angle
27) I am now thinking that there is an homotecia between KLM and ABC. And O is the center. This is because the quadrilateral KLCA is an isosceles trapece and O is the crosspoint of it´s diagonals. And then, O is the baricenter from the ABC, which is equilateral too.
28) So, BO is the double from BM. And, since BM belongs to the mediana from the triangel SRM, then O is the middle point from the segment MG, where G is the baricenter from the triangle RSM.

I think that we are pretty close now.

29) The line BM is divided in 3 by O and by G. The two triangles ABC and RSM are equilateral with the same mediana, so they are equals to each other. The set of the two of them looks like the jewish stern.

30) Let 1 be the side from RMS. So, BM is (sqrt3)/2 and BO is 2/3 of this measure. BM is then 1/sqrt3.

31) You have the sides BR and BO from the right triangle RBO, it's 1/2

tan ROB = opposite side/adjacent side = (1/2)/(1/sqrt3) = sqrt 3/2, which is less than 1, so the angle is less than 45. And the double of an angle that is less than 45o is acute

Wow!! I bet that there are much better ways to prove this. But this is at least something, and, considering that it´s 5 am, it´s a miracle that I could solve this complicated problem. Well, at least, I hope I solved it and I haven´t done any inaccuracies of false statements.

Take care

Ana

Answer 2

I am not sure that I have the right picture, but if I do and the definition of acute is less than 90 degrees. I think that I can draw a your figure with ROS being 90 degrees. Can you? I assume that ABC is any triangle not just equilateral?

Answer 3

I just emailed you. Send me the diagram and I will see if I can help you tomorrow. I am too tired to think of Metric Geometry without a diagram in front of the eyes

Anabel