We have to show that $\frac r q = \frac{r_1}{q_1} \cdot \frac{r_2}{q_2}$. Let $s, s_1, s_2$ be the semiperimeters, $\triangle, \triangle_1, \triangle_2$ the areas and h the common C-altitude of the triangles $\triangle ABC, \triangle AMC, \triangle BMC$. The circle pairs $(r, q), (r_1, q_1), (r_2, q_2)$ are centrally similar with the common external homothety center C, the intersection of their common external tangents. Their pairwise homothety coefficients are equal to the distances of their tangency points on from the vertex C: $\frac r q = \frac{s - AB}{s},\ \ \frac{r_1}{q_1} = \frac{s_1 - AM}{s_1},\ \ \frac{r_2}{q_2} = \frac{s_2 - BM}{s_2}$ and we have to show that $(?)\ \ \ \frac{s - AB}{s} = \frac{s_1 - AM}{s_1} \cdot \frac{s_2 - BM}{s_2}$ $(?)\ \ \ 1 - \frac{AB}{s} = \left(1 - \frac{AM}{s_1}\right) \left(1 - \frac{BM}{s_2}\right)$ $(?)\ \ \ \frac{AB}{s} = \frac{AM}{s_1} + \frac{BM}{s_2} - \frac{AM}{s_1} \cdot \frac{BM}{s_2}$ Substituting $AB = \frac{2 \triangle}{h},\ AM = \frac{2 \triangle_1}{h},\ BM = \frac{2 \triangle_2}{h}$ and $s = \frac{\triangle}{r},\ s_1 = \frac{\triangle_1}{r_1},\ s_2 = \frac{\triangle_2}{r_2}$, this becomes $(?)\ \ \ \frac{2r}{h} = \frac{2r_1}{h} + \frac{2r_2}{h} - \frac{4r_1 r_2}{h^2}$ Reducing by $\frac{2h}{r_1 r_2}$ and rearranging, $(?)\ \ \ \frac{2}{h} = \frac{1}{r_1} + \frac{1}{r_2} - \frac{r}{r_1 r_2}$ This relation has been proved in the Inradii problem and it keeps popping up - I already used it to solve the easy problem Congruent inscribed circles and not so easy problems A very difficult one and Finding ratio. Yetti

It is maybe not so easy to see that all 6 radii in the expression q*r1*r2 = r*q1*q2 are segments of the sides of a triangle in which concurrent cevians are approprietly cutting the sides into. Ceva theorem works just fine in this particular case. Some may have preferrence for Menelaos theorem. It will produce the desired result too. In the case of Ceva, any idea what the point of concurrency may be? Thank you. M.T.

armpist wrote: It is maybe not so easy to see that all 6 radii in the expression q*r1*r2 = r*q1*q2 are segments of the sides of a triangle in which concurrent cevians are approprietly cutting the sides into. Ceva theorem works just fine in this particular case. Some may have preferrence for Menelaos theorem. It will produce the desired result too. In the case of Ceva, any idea what the point of concurrency may be? Thank you. M.T. It is quite easy to see that a triangle does exist, the sides of which can be cut in any ratios whatsoever. On the other hand, it is also quite easy to find a triangle and a point M on one of its sides, such as q + r > q > q1 + r1 + q2 + r2, which means that triangles with the sides q + r, q1 + r1, q2 + r2 or q, q1, q2 cannot exit. So, it has to be a triangle with the sides r, r1, r2, which trivially exists, because r1 + r2 = r + 2*r1*r2/h > r. But the thingy is this: To use Ceva's theorem to deduce q*r1*r2 = r*q1*q2, you have to know the concurrency point to begin with, don't you?

For a change, here are my questions, instead of solution: Any way of easy construction of the triangle for a proof without words? How does this radii triangle gets transformed as point M moves along side AB? At some points the break-up of the sides changes: if a side was (q + q1) then it changes to (q + q2) (and appropriate change for other sides) What is the path of the cevian concurrency point? When does it change it's position from inside to the outside of this triangle with sides divided in proportions of 6 radii? Any connection with Ph-An's favorite problem of equal length cevians intersecting in circumscribed quadrilateral (that you Yetti solved recently)? You must have noticed that the 4 centers r1,r2,q1,q2 are concyclic when ABC is isosceles ! That's what we should have looked at then. Any 'after the fact' beneficial use of the present identity in the old problem? Any connection to axial inversion? The identity r / q = r1 / q1 * r2 / q2 r / q - constant and r1 /q1 * r2 / q2 look very much like tan(X) * tan(Y). One more thing I forgot to tell you: the configuration in Inradii problem works for a simple reason - if you cross-connect ex-centers with their ex-circles tangency points they will intersect on altitude right in the middle of it, and not because of complicated multiplication and/or division. There are much stronger forces, stronger than arithmetic operators, something like projectivity (or collinear UFOs) ! Thank you. M.T.

Take a look at this seemingly related topic http://mathworld.wolfram.com/CrossedLaddersTheorem.html in particular H. Stengel contribution. Thank you. M.T.

armpist wrote: Any 'after the fact' beneficial use of the present identity in the old problem? Any connection to axial inversion? The identity r / q = r1 / q1 * r2 / q2 r / q - constant and r1 /q1 * r2 / q2 look very much like tan(X) * tan(Y) As for the 1st question, I already gave 3 examples. How can you pose a question that has been answered shortly before you even asked it? As for the following expressions, you are welcome to use some multiplications, divisions, etc. armpist wrote: One more thing I forgot to tell you: the configuration in Inradii problem works for a simple reason - if you cross-connect ex-centers with their ex-circles tangency points they will intersect on altitude right in the middle of it, and not because of complicated multiplication and/or division. There are much stronger forces, stronger than arithmetic operators, something like projectivity (or collinear UFOs) ! Thank you. M.T. That's interesting. Can you prove it? It could well be called Armpist's paradox, because it is patently false (just draw the figure). What makes simple and even not so simple mathematical apparatus useful is the fact that it is rigged by axioms to reflect reality. If the rigging is right, the "forces" are contained in it.

armpist wrote: Take a look at this seemingly related topic http://mathworld.wolfram.com/CrossedLaddersTheorem.html in particular H. Stengel contribution. Thank you. M.T. Just leave me alone, will you? I really do not want to talk to you, but it requires some effort on your side as well.

Here: $\frac{1}{r}-\frac{1}{q}=\frac{1}{r_{1}}-\frac{1}{q_{1}}=\frac{1}{r_{2}}-\frac{1}{q_{2}}$

$\frac{1}{rq_{1}}+\frac{1}{r_{1}q_{2}}+\frac{1}{r_{2}q}=\frac{1}{rq_{2}}+\frac{1}{r_{1}q}+\frac{1}{r_{2}q_{1}}$