Prove that $SD'$, $EF$and $HI$ are concurrent?

Let $$\triangle ABC$$ be a triangle with incenter $$I$$ and orthocenter $$H$$. The incircle of $$\triangle ABC$$ touches $$BC$$, $$CA$$, $$AB$$ at $$D, E, F$$ respectively. Let $$D'$$ be the reflection of $$D$$ through $$I$$ and let $$S$$ be the midpoint of $$AI$$. How can we prove that $$SD'$$, $$EF$$and $$HI$$ are concurrent?

• I can bash this in a reasonable length though I would like to see an elegant solution. – dezdichado Oct 18 '18 at 17:20

I'll prove the statement using complex coordinates so I'll denote all points with small letters, sorry. I won't elaborate every single detail but you should still be able to follow the proof.

Put the triangle into a complex plane and, WLOG, assume that the radius of the inscribed circle is equal to 1. Choose point $$i$$ as the origin, with real axis along the line $$id$$ (see picture).

Draw point $$v$$ symmetric to $$d'$$ with respect to $$s$$. Because $$as=si$$, $$sd'=sv$$ and $$\angle asv=\angle isd'$$ triangles $$\triangle asv$$ and $$\triangle isd'$$ are congruent (SAS). Consequentially $$av=id'=1$$

As a consequence $$\angle sav=\angle sid'$$ so lines $$av$$ and $$id'$$ must be parallel. Line $$d'i$$ is perpendicular to $$bc$$ so line $$av$$ must be perpendicular to $$bc$$ too, thus passing through orthocenter $$h$$.

Taking all that into account, it means that triangles $$\triangle ud'i$$ and $$\triangle uvh$$ are similar.

To finish the proof we need two lemmas and I won't prove them in detail:

Lemma 1:

$$a=\frac{2ef}{e+f}$$

You can find the proof here on page 5. Even without a proof you should be able to figure it out by yourself.

Lemma 2:

$$h=\frac{2(d^2e^2+e^2f^2+f^2d^2+def(d+e+f))}{(d+e)(e+f)(f+d)}$$

The proof can be found here, on page 4.

So we have the following coordinates:

$$i=0\tag{1}$$ $$d'=-1\tag{2}$$ $$v=a+1=\frac{2ef}{e+f}+1\tag{3}$$ $$h=\frac{2(d^2e^2+e^2f^2+f^2d^2+def(d+e+f))}{(d+e)(e+f)(f+d)}\tag{4}$$

Because triangles $$\triangle ud'i$$ and $$\triangle uvh$$ are similar:

$$\frac{v-u}{d'-u}=\frac{h-u}{i-u}\tag{5}$$

Replace (1), (2), (3), (4) into (5) and solve for $$u$$:

$$u=\frac{e^2 f^2+e^2 f+e^2+e f^2+e f+f^2}{e^2 f+e f^2+2 e f+e+f}\tag{6}$$

The points $$u$$, $$e$$ and $$f$$ are collinear if and only if:

$$\lambda=\frac{u-e}{f-u}\in R\tag{7}$$

...or, if you replace (6) into (7):

$$\lambda=\frac{(e+1) f \left(e^2-f\right)}{e (f+1) \left(e-f^2\right)}\in R\tag{8}$$

The trick is to prove that (8) is a real number for any $$e,f$$ from the unit circle. This is equivalent to proving:

$$\lambda - \bar\lambda=0$$

or:

$$\frac{(e+1) f \left(e^2-f\right)}{e (f+1) \left(e-f^2\right)} - \frac{(\bar e+1) \bar f \left(\bar e^2-\bar f\right)}{\bar e (\bar f+1) \left(\bar e-\bar f^2\right)}=0$$

$$e^3 \bar{e}^2 f \bar{f}+e^3 \bar{e}^2 f+e^3 (-\bar{e}) f \bar{f}^3-e^3 \bar{e} f \bar{f}^2-e^2 \bar{e}^3 f \bar{f}\\ -e^2 \bar{e}^3 \bar{f}+e^2 \bar{e}^2 f-e^2 \bar{e}^2 \bar{f}-e^2 \bar{e} f \bar{f}^3+e^2 \bar{e} \bar{f}^2\\ +e^2 f \bar{f}^2+e^2 \bar{f}^2+e \bar{e}^3 f^3 \bar{f}+e \bar{e}^3 f^2 \bar{f}+e \bar{e}^2 f^3 \bar{f}-e \bar{e}^2 f^2\\ -e \bar{e} f^3 \bar{f}^2+e \bar{e} f^2 \bar{f}^3-e f^3 \bar{f}^2-e f^2 \bar{f}^2-\bar{e}^2 f^2 \bar{f}-\bar{e}^2 f^2\\ +\bar{e} f^2 \bar{f}^3+\bar{e} f^2 \bar{f}^2=0\tag{9}$$
It looks pretty hopeless, but for points $$e$$ and $$f$$ on the unit circle:
$$e\bar e = f \bar f =1\tag{10}$$
Replace (10) into (9) and you will prove that (9) is indeed zero. In other words $$\lambda - \bar\lambda=0$$ for any $$e,f$$ so $$\lambda$$ must be a real number. Therefore points $$e,f,u$$ are collinear and lines $$sd'$$, $$hi$$ amd $$ef$$ are congruent.