While the article The Centroid of A Triangle explains how to find the centroid of any triangle, this article elaborates on how to determine the centroids of some special-typed triangles, that is isosceles, equilateral, and right-angled triangles.

**The Centroid of An Isosceles Triangle**

Consider the triangle ABC below. In this triangle, AC = BC. Let D and E be the midpoints of the line segments **BC** and **AB**, respectively.

**Figure 1**

Since ΔABC is isosceles with AC = BC, the median **CE** is an altitude of ΔABC as well. So, CE ⊥ AB. Let AC = b, BC = a and AB = c. Then, by the Pythagorean formula, it can be shown that the altitude CE has the length of t_{C} satisfying:

In the article The Centroid of A Triangle, it has been proven that CT : TE = 2 : 1. As a consequence, , and it follows that

If we define x_{C} = AE and and y_{C} = TE then: (See Figure 2 below)

If T’ is the foot of the perpendicular from T to the side **BC**, what are CT’ and TT’? (See Figure 2)

**Figure 2**

Assuming ∠T’CT = θ, we have CT’ = CT cos θ. In ΔCEB, whereas . Hence . By (*), it can be concluded that .

In the article The Centroid of A Triangle, it has been proven that the distance between the centroid of a triangle and the base of the triangle is one third of the length of the altitude drawn from the vertex opposite to the base. Therefore, where t_{A} is the length of the altitude drawn from A to the side **BC**.

From the ratios t_{A} : t_{C} = AB : BC = c : a and the fact that , as given by (*), it follows that . In addition, . If we let CT’ = x_{A} and TT’ = y_{A}, it can be concluded that:

The projection of T on the side **AC** side gives similar results. Let T” be the foot of the perpendicular from T to the side **AC**. Since ΔABC is symmetrical about the line segment **CE**, it holds that CT” = CT’ and TT” = TT’. If we let CT” = x_{B} and TT” = y_{B}, we have:

As *a* = *b* in the isosceles triangle ABC, the formulae x_{B} and y_{B} can be easily obtained from x_{A} and y_{A}, i.e. by replacing *a* with *b*, and vice versa.

**The Centroid of An Equilateral Triangle**

Equilateral triangles are also isosceles triangles. Consequently, all the formulae derived above apply to equilateral triangles as well. Suppose that ABC is an equilateral triangle. Therefore, a = b = c, where a = BC, b = AC, and c = AB. (See Figure 3 below.) Replace *c* with *a* in the x_{A} and y_{A} formulae above. This results in:

**Figure 3**

Since *a* = *b* = *c*, and .

**The Centroid of A Right-Angled Triangle**

Consider a right-angled triangle ABC with **AC** ⊥ **AB**. Let AB = c, AC = b, and BC = a. Place the triangle on the Cartesian plane so that **AB** coincides with the x-axis, **AC** coincides with the y-axis. (See Figure 4 below.)

**Figure 4**

It follows that A(0,0), B(*c*,0), and C(0,*b*). Referring to the article The Centroid of A Triangle, the coordinates of the centroid of the triangle, that is T, can be determined by the formula:

Let , and , it can be concluded that .