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I am struggling to show that constructed triangle is indeed the desired one

Condition: In triangle $ABC$, one has marked the in-center, the foot of altitude from vertex $C$ and the center of the ex-circle tangent to side $AB$. After this, the triangle was erased. Restore it.

Analysis: Let $I$ be the in-center and $I_C$ be the ex-center and $H$ be the foot of altitude from $C$.

There are several ways of showing that $AB$ is the angle bisector of $\angle IHI_C$. This would restore the side $AB$. Lines $II_A$ and perpendicular to $H$ on $AB$ would intersect at point $C$.

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Since outer and inner angle bisector are perpendicular we see that $A$ and $B$ are on circle with diameter $I_C I$. Then angle bisector of $IHI_C$ cuts this circle at $A$ and $B$. Now the perpendicular through $H$ on $AB$ cuts line $I_C I$ at $C$ and we are done.

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  • $\begingroup$ Nice solution. Can you explain why $HI$ and $HI_C$ are symmetric with respect to $AB$? $\endgroup$ – Jack D'Aurizio Mar 28 '18 at 15:34
  • $\begingroup$ I don't know yet, I used his hint. $\endgroup$ – Aqua Mar 28 '18 at 15:38
  • $\begingroup$ I get this part but how can I be sure that I and I-C are indeed the in-center and ex-center respectively from this construction? $\endgroup$ – Andrew Mar 28 '18 at 18:24
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Just a little addendum to ChristianF's answer, explaining why $AB$ bisects $\widehat{IHI_C}$.
Let $J$ and $K$ be the projections of $I$ and $I_C$ on $AB$.

enter image description here

We have $AJ=BK=\frac{b+c-a}{2}$ and $AH=b\cos A=\frac{b^2+c^2-a^2}{2c}$, so $HJ=AJ-AH=\frac{(a-b)(a+b-c)}{2c}$.
Similarly $HK=AK-AH=\frac{(a-b)(a+b+c)}{2c}$. Since $r=IJ=\frac{2\Delta}{a+b+c}$ and $r_C=\frac{2\Delta}{a+b-c}$ we have

$$ \frac{r_c}{HK}=\frac{4c\Delta}{(a-b)(a+b-c)(a+b+c)}=\frac{r}{HJ}$$ hence $AB$ bisects $\widehat{IHI_C}$.

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  • $\begingroup$ Great!.......... $\endgroup$ – Aqua Mar 28 '18 at 18:06

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