## The point of intersection of the heights, angle bisectors and medians of a regular triangle divides them into two line segments; the length of the longer one of these line segments from is equal to the length of the circumscribed circle’s radius and the length of the shorter one to the length of the inscribed circle’s radius: where:

h – the heights of a regular triangle;

m – the medians of a regular triangle;

l – the angle bisectors of a regular triangle;

k – the perpendicular bisectors of a regular triangle. The medians, heights, angle bisectors and perpendicular bisectors with the circumscribed circle’s radius and inscribed circle’s radius of a regular triangle

In other words,

In a regular triangle the distance from the point of intersection of the heights, angle bisectors and medians

• ## to any vertex is equal to the length of the circumscribed circle’s radius. The distance from the point of intersection of the heights, angle bisectors and medians to any vertex

In a regular triangle the distance from the point of intersection of the heights, angle bisectors and medians

• ## to any side of a triangle is equal to the length of the inscribed circle’s radius. The distance from the point of intersection of the heights, angle bisectors and medians to any side of a triangle

## Step 1

Consider equilateral triangle ABC (АВ=ВС=АС).

Let BF, AD, CE be the medians.

According to the property of the theorem on the median of a regular triangle, in a regular triangle, the median drawn to any side is its angle bisector, height and perpendicular bisector. Proof of the theorem. Step 1

## Step 2

Medians of a triangle are divided by the point of intersection in the ratio of 2:1 counting from the vertex.

Let O denote the intersection of the medians.

Then:  Proof of the theorem. Step 2

## Step 3

The lengths of the medians of a regular triangle are equal (CE=AD=BF).

From step 2: Therefore:  Proof of the theorem. Step 3

## Step 4

The inscribed circle’s center of a triangle lies at the intersection of the angle bisectors.

О – is is the point of intersection of the angle bisectors of the triangle by condition. Therefore O is the inscribed circle’s center of a triangle.

The distances from the point O to any side of a triangle are equal.

Therefore, these distances will be the radii of the inscribed circle:  Proof of the theorem. Step 4

## Step 5

As: And: Then:  Proof of the theorem. Step 5

## Step 6

The circumscribed circle’s center of a triangle lies at the intersection of the perpendicular bisectors.

Point O is is the point of intersection of the perpendicular bisectors of the triangle.

Therefore O is the circumscribed circle’s center of a triangle.

So, the distance from the point O to any vertex is the length of the circumscribed circle’s radius.  Proof of the theorem. Step 6

## Step 7

СЕ = СО+ОЕ

СО = R (step 6)

ОЕ = r (step 4)

Therefore: According to the property of the regular triangle, all heights, medians, angle bisectors and perpendicular bisectors of a regular triangle are equal in length, then their length will be CE, and, therefore, their length will be the sum of the radii of the inscribed and circumscribed circles. The theorem is proved. Proof of the theorem. Step 7