Show that the circumscribed circle for a triangle passes through the middle of the segment determined by the center of the incircle and the center of an excircle.

I found this Incenter and circumcenter of triangle ABC collinear with orthocenter of MNP, tangency points of incircle, but it was just tangentially helpful, and also everything about the Euler line and https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle. Any help, please?

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    $\begingroup$ I believe, you've somewhat complicated your wikipedia search because of odd word "center" in your question. Correct formulation: "Circumscribed circle passes through the middle of the segment determined by the center of the incircle and the center of an excircle". $\endgroup$ Mar 23 at 12:25
  • $\begingroup$ @IvanKaznacheyeu thanks!!! $\endgroup$
    – user318394
    Mar 23 at 12:38
  • $\begingroup$ @AnatolyDenicula: I have rolled the question back to a state where it actually contained an exercise, so that the answers you've received remain relevant. (Please note that dramatically altering the nature of a question —even your own— may be considered vandalism, which is against community guidelines. Besides, making answers appear irrelevant is inconsiderate to the answerers who invested time and effort in helping you.) $\endgroup$
    – Blue
    Apr 14 at 5:19

2 Answers 2


Proof. Note that the incenter, $I$ is determined by the intersection of the internal angle bisectors of the triangle. We will consider the $A$-excircle, i.e. the excircle whose center, let's call it $X$, is determined by the intersection of the external angle bisectors at vertices $B$ and $C$. Note that $X$ lies on the line $AI$ as well.

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Let $\angle BAC = \alpha$, $\angle ABC = \beta$ and $\angle ACB = \gamma$. Furthermore, let $AX \cap (ABC) = M$, so we need to show that $MI = MX$. We claim that $BICX$ is a cyclic quadrilateral centered at $M$. Firstly, we have $\angle BAM = \angle CAM = \frac{\alpha}{2}$, so $MB = MC$. Next, we show that $\triangle MCI$ is isosceles. Indeed, we have $\angle MIC = \angle IAC + \angle ACI = \frac{\alpha}{2} + \frac{\gamma}{2} = \angle BAM + \angle BCI = \angle BCM + \angle BCI = \angle MCI$.

Since $MB = MC = MI$, $M$ is the circumcenter of $\triangle BCI$, it remains to show that $X ∈ (BCI)$. We have $\angle BIC = 180° - \frac{\beta}{2} - \frac{\gamma}{2}$. On the other hand, $\angle BXC = 180° - \angle CBX - \angle BCX = 180° - \frac{180° - \beta}{2} - \frac{180° - \gamma}{2} = \frac{\beta}{2} + \frac{\gamma}{2}$.

Therefore, $\angle BIC + \angle BXC = 180°$, i.e. $X ∈ (BCI)$ and $(BCI)$ is centered at $M$, so $MI = MX$, as desired.


Let $I$ is incircle center, $O$ is circumscribed circle center, $X$ is center of excircle that is tangent to edge $AB$, $P$ is middle of $IX$ segment.

$\angle IAB=\frac{\angle CAB}{2}$, $\angle XAB=\frac{180°-\angle CAB}{2}=90°-\angle IAB$, $\angle XAI=\angle IAB+\angle XAB=90°$. In the same way, $\angle XBI=90°$. Then $XAIB$ is cyclic quadrilateral with diameter $XI$ and center $P$, then $PA=PB$. Then $P$ is on perpendicular bisector of $AB$, as well as $O$. Then $OP$ is perpendicular bisector of $AB$.

$\angle AOP=\frac{\angle AOB}{2}=\angle ACB$. $\angle APO=\frac{\angle APB}{2}=\angle AXB=$ $180°-\angle AIB=\angle IAB+\angle IBA=$ $\frac{\angle CAB}{2}+\frac{\angle CBA}{2}=$ $\frac{180°-\angle ACB}{2}=90°-\frac{\angle ACB}{2}$. $\angle PAO=180°-\angle AOP-\angle APO=$ $180°-\angle ACB-90°+\frac{\angle ACB}{2}=$ $90°-\frac{\angle ACB}{2}=\angle APO$, therefore $OA=OP$. OK.


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