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.

  • $\begingroup$ 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

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 $\widehat{IHI_C}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.