# The theorem on the angle bisector of a regular triangle

In a regular triangle, the angle bisector drawn to any side is its height, median and perpendicular bisector. The theorem on the angle bisector of a regular triangle

## Step 1

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

Let BF, AD, CE be the angle bisectors.

We will prove that they are the heights, medians and perpendicular bisectors. Proof of the theorem on the angle bisector. Step 1

## Step 2

As АВ=АС and AD is the angle bisector to base BC, then according to the property of an isosceles triangle, AD is the median and the height.

As AD is perpendicular to side BC and divides it in half, then AD is the perpendicular bisector. Proof of the theorem on the angle bisector. Step 2

## Step 3

As АС=ВС and СЕ is the angle bisector to base AB, then according to the property of an isosceles triangle, CE is the median and the height.

As CE is perpendicular to side AB and divides it in half, then CE is the perpendicular bisector. Proof of the theorem on the angle bisector. Step 3

## Step 4

As АВ=ВС and BF is the angle bisector to base AC, then according to the property of an isosceles triangle, BF is the median and the height.

As BF is perpendicular to side AC and divides it in half, then BF is the perpendicular bisector.

The theorem is proved. Proof of the theorem on the angle bisector. Step 4