**Yes, you are reading this correctly: a triangle can have four sides. If this sounds strange, I have to agree: this statement could be formulated in a more nuanced way: when we look from the point of view of the associated circles to the sides of a triangle, the fourth side arises. **

These circles related to a triangle are the circumcircle, incircle and the nine-point circle. The properties of these circles, and what they look like, is background information that may help to form the line of reasoning for this article.

The simplest circle could be understood by its name, the circumcircle, whereby all the corners of the triangle are on this circle. The matching center of this circle is called the circumcenter. To construct this circle, given a triangle, you create on two sides the perpendicular bisector (it has three, but drawing two is enough). Where those lines intersect, is the circumcenter, called *O*. This works because, when you construct all those perpendicular bisectors, you create six right-angled triangles, all with the same length for the hypotenuse, the radius.

The second circle mentioned is the incircle. This circle touches all sides of the triangle from the inside. To construct this circle, make two angle bisectors, and where those lines meet, you have the incentre,* I*. From the incentre to one side is the radius, inradius r, of the incircle. Quickly said, the reason this works, is that the similar triangles are created if perpendiculars are drawn from *I *to the sides and so the distances from *I *to the three sides are equal.

Furthermore, we have the nine-point circle. This circle passes through nine definite points of the triangle: all the midpoints of the sides, the intersections between the altitudes with the sides, and through the midst of the distance between the orthocenter (the intersection between all the altitudes, named *H*) and the angles, thus nine points in total. To form this circle, make another triangle connecting all the midpoints of the sides of your initial triangle. Then, create two perpendicular bisectors. Where these cross, we have the center of our nine-point circle. The distance between that center and one of the nine-points is the radius of the nine-point circle. Click* here* for a proof for the nine-point circle, as this one is a little bit too long to mention.

From the properties of these three circles it follows that for most triangles a fourth side arises. Concisely, all sides of the triangle, including the fourth, must meet the following requirements:

a. Each side has its endpoints on the circumcircle;

b. Each side has its midpoints on the nine-point circle; and

c. Each side touches the incircle

That being said, how do you find such a fourth side that all these conditions are met? We have to follow the main principle: there are three criteria for the three circles in order to be in those relationships (a, b and c) to the sides of a triangle and the circles:

1. The incircle is inside both the circumcircle and the nine-point circle, and touches the nine-point circle;

2. Let the circumradius be R, the inradius r, and the distance between the incentre *I *and the circumcentre *O *be d, then:

3. The radius of the nine-point circle is` `

Earlier I explained how to construct the circle when a triangle is given, but with this information it is possible to create the three circles if you have the three center points only.

To find the fourth side, we have to use a certain technique that lets the fourth side satisfy conditions a, b and c. First, I am going to tell you what you have to do to get the fourth side. Then an explanation follows. Draw a triangle to your preference. Next create all three corresponding circles. Get a ruler, and mark a point *c* as the middle point, then mark two other points on the ruler with a distance r (inradius) from *c*. After that, make another circle with diameter *OI, *and place *c* somewhere on the line of this circle. Now you can move point *c *over the circle. Let the other edge of the ruler always pass *O. *In this way the middle mark traces out the circle, until one of the marks (not c) passes the nine-point circle. Mark this point and go further with the procedure. If we are facing a regular triangle, not an isosceles triangle, we get *four *points, A, B, C and* M, *on the lines which satisfy the conditions a, b and c.

This technique works due to the fact that a rectangle *PITM *is created (we use the same *I* as mentioned before). The rectangle is constructed by drawing a line through *O *and our new point, *M*, where *P* is on this line and is connected to *I *in a right angle*. M* and T are also connected. It makes a rectangle because we have used r as the distance from *P* to *M *and as you might know, the radius of the circle is perpendicular to the tangent line, in this case one of our four sides. To make things clear, this technique works as we found a way that this rectangle is correctly in correspondence to our points (center points) and our triangle;

i. *P *is on the line *OM;*

ii. *PM *= r;

iii. ⦣ *OPI *= 90 °

As brought up earlier, this cannot be the case for isosceles triangles, since the nine-point touches the incircle. The fourth side coincides with the base.

Finally, a fourth side may seem a little bit less silly now. Or not… that is also possible, nevertheless, from an analytic perspective, that from the descriptive circles of a triangle there is indeed a fourth side (and thus not for isosceles triangles). If you’re not convinced yet,* h**ere* is the proof.

**Source: **