1. ## triangle question

Hi, here is my problem:

Inside of a triangle $ABC$, a random point $M$, in which three straight lines parallel to the respective sides of the triangle intersect, was chosen. Those straight lines have divided the triangle into six parts, three of which are also triangles. Let $r_1, r_2, r_3$ be the radii of circles inscribed in those three triangles, and let $r$ be the radius of a circle inscribed in triangle $ABC$. Show that $r=r_1+r_2+r_3$.

Help appreciated. Thank you.

2. I found it !

Okay look at the sketch. It is not necessary to draw the inscribed circles.

Consider triangles DEM and ABC. All their sides are parallel to each other. Hence they're congruent (or similar). Let $k_1$ be the ratio between the two. This means that $k_1=\frac{DE}{AB}=\frac{DM}{AC}={\color{red}\frac{ EM}{BC}}={\color{blue}\frac{r_1}{r}}=\dots$
This always holds true for two congruent triangles.

It'll be the same for triangle FGM : $k_2={\color{blue}\frac{r_2}{r}}={\color{red}\frac{ MG}{BC}}=\dots$

and for triangle HMI : $k_3={\color{blue}\frac{r_3}{r}}={\color{red}\frac{ HI}{BC}}$

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Now consider the blue parts of the equalities.
You can see that if you sum them, it should give 1, according to the result you're given.

So we'll prove that the sum of the red parts is equal to 1 !

${\color{red}\frac{EM}{BC}+\frac{MG}{BC}+\frac{HI}{ BC}}$

But since EM//BH and EB//MH (parallel), then EMHB is a parallelogram. Hence $\bold{EM=BH}$

Similarly, MGCI is a parallelogram. And hence $\bold{MG=IC}$

So the red sum becomes :
$=\frac{BH}{BC}+\frac{IC}{BC}+\frac{HI}{BC}=\frac{B H+IC+HI}{BC}$

you can see that the points B, H, I and C are aligned. Hence $BH+IC+HI=BC$

So the red part is equal to 1.

Therefore for the blue part :
$1=\frac{EM}{BC}+\frac{MG}{BC}+\frac{HI}{BC}=\frac{ r_1}{r}+\frac{r_2}{r}+\frac{r_3}{r}=\frac{r_1+r_2+ r_3}{r}$

And you're done