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.

Медиана равностороннего треугольника

The theorem on the median of a regular triangle

Proof of the theorem on the median

Step 1

 

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

Let BF, AD, CE be the medians.

We will prove that they are the angle bisectors, heights and perpendicular bisectors.

Медиана равностороннего треугольника

Proof of the theorem on the median. Step 1

Step 2

 

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

As AD is perpendicular to side BC and divides it in half, then AD is the perpendicular bisector.

Медиана равностороннего треугольника

Proof of the theorem on the median. Step 2

Step 3

 

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

As CE is perpendicular to side AB and divides it in half, then CE is the perpendicular bisector.

Медиана равностороннего треугольника

Proof of the theorem on the median. Step 3

Step 4

 

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

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

 

The theorem is proved.

Медиана равностороннего треугольника

Proof of the theorem on the median. Step 4