# Triangle Construction By In-Center, Ex-Center and foot of an Altitude

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 $C$.

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.

• Nice solution. Can you explain why $HI$ and $HI_C$ are symmetric with respect to $AB$? – Jack D'Aurizio Mar 28 '18 at 15:34
• I don't know yet, I used his hint. – Aqua Mar 28 '18 at 15:38
• 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? – Andrew Mar 28 '18 at 18:24

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 $AB$.

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 $\widehat{IHI_C}$.

• Great!.......... – Aqua Mar 28 '18 at 18:06