By Dario Gonzalez Martinez



A triangle has several interesting properties and relations with other shapes.  For this writing-up we will discuss relations and properties of three important circles associated with a triangle: the incircle, the excircles and the nine-point circle.  First, we need to remember about the definitions and constructions for each of these important circles.





It is the most popular of the mentioned circles that we will discuss.  The incircle is the largest circle contained by a triangle, and it is tangent to each triangle’s sides.  Since this circle is tangent t to each triangle’s side, we can conclude that its center is an equidistant point from the triangle’s sides.


An angle bisector is the locus of all points that are equidistant from two sides of a triangle.  So, an equidistant point from every triangle’s side should lie on each of the three angle bisectors of a triangle, that is, the point lies on the intersection among them.  So, a construction for the incenter could be as follow (figure 1 shows the construction):


Given a triangle ABC


1) Construct the angle bisectors though A and B.

2) I will be the intersection between the angle bisectors.

3) Construct a line through I and perpendicular to one of the triangle’s sides.

4) M will be the intersection between the perpendicular line and the triangle’s side.

5) Construct a circle with center I and radius IM.


Figure 1


The line MI is constructed such that it is perpendicular to AB because segment MI represents the distance from I to side AB, and given that the incircle is tangent to AB, MI and AB must be perpendicular to each other since MI is radius.





These circles related to a triangle are not as known as the incircle, at least, in the school mathematics.  An excircle is a circle tangent to one triangle’s side and the extension of the other triangle’s sides.  In this manner and similarly to the incircle, the center of an excircle (called excenter) is equidistant from one triangle’s side and the extension of the other two.


According to the definition above, we could find an excenter by constructing the external angle bisector and locate the intersection point between them.  Therefore, a construction for an excircle could be the following:


Given a triangle ABC


1) Extend sides AB and CB in the direction opposite their common vertex.

2) Construct the angle bisectors for the external angles at A and C.

3) J is the intersection between the angle bisectors.

4) Construct a line though J and perpendicular to one of the extensions.

5) D is the intersection between the perpendicular line and the extension.

6) Construct a circle with center J and radius JD.


Figure 2 below shows the construction mentioned:


Figure 2


Of course, there are three excircles since a triangle has three sides.  Actually, if we construct the three excircles, we obtain an image as follow:


Figure 3





I. Simple but interesting properties


We can conclude some important properties and relations if we construct the incircle and excircles of a triangle simultaneously.  Consider figure 4 below which shows the incircle and excircles of a triangle ABC:


Figure 4


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One important fact is that J, C and K are collinear points.  We know that angles ACD and BCL are equal to each other since they are opposite angles.  Therefore, angle bisector CJ and CK are the same bisector, so point J, C and K are collinear.  We can elaborate similar reasoning to prove that each trio J, A and M, and M, B and K are also collinear, respectively.  Thus, triangle JMK inscribes triangle ABC.


Moreover, we can prove that points C, I and M are collinear.  We know that C and I lie in the angle bisector of the angle ACB.  M is an excenter, so it is equidistant from the extension of the sides CA and CB, which are also the extensions of the angle ACB’s sides.  Therefore, M must lie on the angle bisector of the angle ACB, and then C, I and M are collinear.  We can elaborate similar reasoning to prove that each trio A, I and K, and B, I and J are also collinear, respectively.


Another interesting relation is that CM and JK are actually perpendicular to each other (and for similar reasons BJ is perpendicular to MK, and KA is perpendicular to JM).  Consider



Since BCD is a straight line, then



Therefore, angle JCI is a right angle, and CM and JK are perpendicular to each other.  We can elaborate similar arguments to prove that BJ is perpendicular to MK, and KA is perpendicular to JM.  So, AK, BJ and CM are the altitudes of the triangle JMK, which means triangle ABC is the Orthic Triangle of triangle JMK, and the incenter I of triangle ABC is the orthocenter of triangle JMK.


II. Relation between triangle’s area and incircle and excircles


There exists an interesting relation between the triangle and its inradius.  We should remember that the incenter is equidistant to each triangle’s side and consider the figure 5 below:


Figure 5


We will denote the area of triangle ABC by A.  Consider IE = IF = IG = r inradius and BC = a, AC = b and AB = c, so



Where s is the semiperimeter.  At the same time, we can write






Every triangle has a nine-point circle.  It is called nine-point circle because this circle passes through 9 significant points of a triangle.  These points are:


·         The midpoint of each triangle’s side.

·         The foot of each triangle’s altitude.

·         The midpoint of the segment from each triangle’s vertex and the orthocenter.


Figure 6


There is an easy way to construct the nine-point circle by considering its definition.  Since the nine-point circle must pass through the foot of each triangle’s altitude, we can consider the nine-point circle as the circumcircle of the orthic triangle of triangle ABC.  The circumcircle is unique since the circumcenter is unique for a determined triangle; therefore, if the nine-point circle circumscribes the orthic triangle of triangle ABC, it must be its circumcircle.


The common proof of the nine-point circle’s existence is a bit long.  However, I found an interesting and accessible proof in the web site Polymatematics.  Here is the link




The proof is based on two facts:


1) A circle is completely determined by three points.

2) A lemma that states that a circle whose diameter is side AB of a triangle contains the feet of the altitudes from A and B.


The lemma is related to the fact that every right angle can be circumscribed in a semi-circumference.  Observe figure 7 below:


Figure 7


We can see that H1 and H2 are actually the foot of the altitudes from A and B, respectively since the angles at H1 and H2 are right because they are inscribed in a semi-circumference.


Now consider the figure 8 below:


Figure 8


Let M, Q and P be the midpoints of the triangle’s sides.  Since MQ is a midline of the triangle, it is parallel to H1P, making quadrilateral MQPH1 a trapezoid.  Moreover, QP is also a midline of the triangle ABC, so it is half the length of AB.  Note that M is the center of the circle (since its diameter was AB), and that makes MH1 a radius of the circle, and therefore half the length of AB.  Therefore, the trapezoid is actually an isosceles trapezoid.


We can see that angles MQP and H1PQ are supplementary since MQ is parallel to H1P.  At the same time, the opposite angles of trapezoid MQPH1 are supplementary due to the trapezoid is isosceles, which makes it a cyclic quadrilateral.  So that means that the circle that contains the three midpoints of the sides also contains H1.  If we repeat this reasoning for all three of the altitude feet, we will conclude that there exists a circle that goes through all six points.


It just remain include the three midpoints of the segments that join the orthocenter ad each triangle’s vertex in the proof.  To that end, we should consider the figure 9 below:


Figure 9



If we let T, Q and R be the midpoints of each side of triangle AHC, then everything that we already discuss for triangle ABC above, it is also true for triangle AHC.  If we consider carefully the altitudes of triangle AHC, we will note that they are the segments AH3, HH2 and CH1.  So, the three feet are H1, H2 and H3.  Thus, the circle we found above is also the circle for the smaller triangle AHC, and it must pass through the midpoints of its sides.  By repeating the previous reasoning, we will prove that the circle also contains S, the midpoint of segment HB.


Thus, every triangle has a circle that passes through nine points: the three midpoints of the sides, the three feet of the altitudes, and the three midpoints of the segments connecting each vertex to the orthocenter.





The nine-point circle has an important relation with the incircle and the excircles of a triangle.  The nine-point circle is tangent internally to the incircle and tangent externally to the excircles.  This theorem is known as the Feuerbach's theorem, and the point where the nine-point circle is tangent to the incircle is called Feuerbach's point.


Figure 10


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