The four main triangle types that result in intersecting lines are shown in figure 1. The four types are centroid, circumcenter, orthocenter, and incenter.
Acute Triangle
An acute triangle is one for which all three angles are less than $180^{\circ}$.
Median
The median of triangle of a triangle is a line or line segment from a vertex to the midpoint of the opposite side.
Centroid
If three medians are constructed from the three vertices, they concur (meet) at a single point. That point is called the centroid. The centroid is a balance point for a triangle because all of the interior triangles that are formed have equal area.
Figure 1: Four triangle types that result in an intersection of three lines. The incenter is found from bisecting all three vertices. The centroid comes from constructing the medians. The orthocenter is caused by dropping a perpendicular from each vertex to the opposite side. The circumcenter is created by drawing perpendiculars to the midpoint of each side.
Circumcenter
The circumcenterof a triangle is a common point that is intersected by all three perpendicular bisectors of the sides. The point does not have to be inside the triangle. If a circle is constructed containing all three points of the triangle, the circumcenter will be the center of the circle.
Orthocenter
The orthocenter is formed by dropping a perpendicular line from a vertex to the opposite side. If the triangle is acute, then the orthocenter will be inside the triangle. The orthocenter is the intersection of a triangle's three altitudes. For a right triangle, the orthocenter will be the vertex of the $90^{\circ}$ angle. Upon constructing four planar points, one of which is an orthocenter for the other three, then we have also created an orthocentric system. The orthocentric system having points, $A,B,C$ and $D$ will have the property that each point is the orthocenter of the other three whenever the remaining three points are formed into a triangle.
Incenter
The incenter is a point constructed by bisecting all three vertices of a triangle. The incenter will always be equidistant from the three sides and thus serves as the center of a circle that is tangent to each side. It is possible to obtain the incenter radius from the formula
$$r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$
It is not difficult to see the proof of this radii formula.
Figure 2: Incenter radius is given by $r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$ . The points of tangency between the triangle sides and the circle divide the perimeter into $6$ segments composed of $3$ equal length pairs. The two segments adjacent to vertex $A$ are equal. The sum of one segment from each pair must be equal to half of the perimeter length, which is $s$ by definition. It follows that the segments adjacent to each vertex $a$ equals to $s$ less the side opposite that vertex. Said another way, since the two segments on side $c$ add up to $c$, it must be that the two segments adjacent to vertex $C$ are each $s-c$.
Heron's formula gives the area of triangle $ABC$ in figure 2 as $Area=\sqrt{s(s-a)(s-b)(s-c)}$. Given $G$ as the center of the incenter circle, and since $r$ represents the height of some sub-triangles, we can easily see that the $3$ triangles made from the vertices and point $G$ have areas $$br/2 =\Delta AGC$$, $$cr/2 =\Delta AGB$$ and $$ar/2 =\Delta CGB$$.
$$Area=\sqrt{s(s-a)(s-b)(s-c)}=\frac{ar}{2}+\frac{br}{2}+\frac{cr}{2}$$
Recalling from Heron's formula that $s=\frac{a+b+c}{2}$, $$Area=\sqrt{s(s-a)(s-b)(s-c)}=s\cdot r$$
Pushing $1/s$ into the radical results in
$$\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=r\qquad\square$$
