## Median of a triangle

**Median:** The line joining the vertex to the midpoint of opposite side is called median of a triangle.

There is one median from each vertex and all three medians are concurrent (i.e. they pass through only one single point) and their point of intersection is called centroid.

## Results and properties on medians of a triangle

### 1. Centroid divides each median in 2 : 1

### Proof

Construction Produced line AD to DF such that GD = DF and join the line BF.

In BDF and CDG

BD = CD ( AD is the median on side BC )

DG = DF ( by construction )

BDF = CDG ( vertically opposite angle )so BDF CDG

therefore BFD = CGD

Since alternate angles are equal , so BF CG or BF CE

Now in ABF

since E is the mid-point of AB and EG BF

Therefore G is the mid-point of AF

so , AG = GF = DG + DF = DG + DG = 2DG ( DG = DF )

=AG : DG = 2 : 1

### 2. Median of a triangle divides the triangle into two equal areas.

### Proof

Construction Draw AE such that AE BC as shown in figure

Area of ABD =

=

Area of ADC =

=

= ( BD = CD )

hence, Area of ABD = Area of ADC

### 3. All the three medians of a triangle divide the triangle into six equal parts.

### Proof

In BOC, OD is the median of triangle

area of BOD = area of COD = yIn AOC, OE is the median of triangle

area of AOE = area of COE = xIn AOB, OF is the median of triangle

area of AOF = area of BOF = zIn ABC, AD is the median of triangle

area of ABD = area of ACD z + z + y = y + x + x 2z = 2x z = x (i)In ABC, BE is the median of triangle

from equaton (i) and (ii)

z = x = y

### 4. ABC is a right angle triangle, right angle at B. If AD and CE are the medians of a triangle, then 4 ( ) = 5

### Proof

In ABD, ABD = + = (1)

Similarly, in BCE, EBC = + = (2)

Adding equ. (1) and (2)

+ + + = + + + + = + + + + = + = + = +4( ) = 5

### 5. If ABC is a triangle in which AD is the median on the side BC, then

### Proof

Draw AEBC

In ABE, by pythagorus theorem

= + (i) = + (ii)Adding equation (i) and (ii)

+ = + + + + = + + 2 (iii)Now in ADE

– =Put this value in equation (iii)

+ = + + 2 – 2 + = + + 2 – 2= + + 2 – 2

= + + 2 – 2

= 2 + 2

= 2( +) (BC = 2BD)

### 6. If in a ABC, AD, BE and CF are the medians then 4

### Proof

Since AD is the median

Therefore (i)

similarly

(ii) (iii)adding (i), (ii) and (iii) ,we get

=2() = 2()

= () =3 () = 4()