# Geometry ( Median of a triangle )

## 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

### Proof

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

In $\triangle$BDF and $\triangle$CDG

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

DG = DF         ( by construction )

$\angle$BDF = $\angle$CDG              ( vertically opposite angle )

so $\triangle$BDF $\cong$ $\triangle$CDG

therefore $\angle$BFD = $\angle$CGD

Since alternate angles are equal , so BF $\parallel$ CG or BF $\parallel$ CE

Now in $\triangle$ABF

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

Therefore G is the mid-point of AF

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

$\frac{AG}{DG}$$\frac{2}{1}$

AG : DG = 2 : 1

### Proof

Construction Draw AE such that AE $\perp$ BC as shown in figure

Area of $\triangle$ABD = $\frac{1}{2}\times base\times height$

$\frac{1}{2}\times AE\times BD$

Area of $\triangle$ADC = $\frac{1}{2}\times base\times height$

$\frac{1}{2}\times AE\times CD$

=  $\frac{1}{2}\times AE\times BD$                 ( BD = CD )

hence, Area of $\triangle$ABD = Area of $\triangle$ADC

### Proof

In $\triangle$BOC, OD is the median of triangle

$\Rightarrow$ area of $\triangle$BOD = area of $\triangle$COD = y

In $\triangle$AOC, OE is the median of triangle

$\Rightarrow$ area of $\triangle$AOE = area of $\triangle$COE = x

In $\triangle$AOB, OF is the median of triangle

$\Rightarrow$ area of $\triangle$AOF = area of $\triangle$BOF = z

In $\triangle$ABC, AD is the median of triangle

$\Rightarrow$ area of $\triangle$ABD = area of $\triangle$ACD

$\Rightarrow$ z + z + y = y + x + x

$\Rightarrow$ 2z = 2x

$\Rightarrow$ z = x                           (i)

In $\triangle$ABC, BE is the median of triangle

$\Rightarrow$ area of $\triangle$ABE = area of $\triangle$CBE

$\Rightarrow$ z + z + x = x + y + y

$\Rightarrow$ z = y                        (ii)

from equaton (i) and (ii)

z = x = y

### Proof

In $\triangle$ABD, $\angle$ABD = $90^{\circ}$ $AB^{2}$ + $BD^{2}$ = $AD^{2}$                         (1)

Similarly, in $\triangle$BCE, $\angle$EBC = $90^{\circ}$ $BC^{2}$ + $BE^{2}$ = $CE^{2}$                             (2)

$AB^{2}$ + $BD^{2}$$BC^{2}$ + $BE^{2}$$AD^{2}$$CE^{2}$ $AB^{2}$ + $(\frac{BC}{2})^{2}$$BC^{2}$ + $(\frac{AB}{2})^{2}$$AD^{2}$$CE^{2}$ $AB^{2}$ + $\frac{BC^{2}}{4}$$BC^{2}$ + $\frac{AB^{2}}{4}$$AD^{2}$$CE^{2}$ $\frac{4AB^{2}+BC^{2}+4BC^{2}+AB^{2}}{4}$ =  $AD^{2}$$CE^{2}$ $\frac{5AB^{2}+5BC^{2}}{4}$ =  $AD^{2}$$CE^{2}$

4($AD^{2}+CE^{2}$ ) = 5$AC^{2}$

### Proof

Draw AE$\perp$BC

In $\triangle$ABE, by pythagorus theorem

$AB^{2}$ = $BE^{2}$ + $AE^{2}$                           (i)

$AC^{2}$ = $CE^{2}$ + $AE^{2}$                           (ii)

$AB^{2}$$AC^{2}$$BE^{2}$ + $AE^{2}$$CE^{2}$ + $AE^{2}$ $AB^{2}$$AC^{2}$$BE^{2}$$CE^{2}$ + 2$AE^{2}$                   (iii)

Now in $\triangle$ADE

$AD^{2}$$DE^{2}$ = $AE^{2}$

Put this value in equation (iii)

$AB^{2}$$AC^{2}$$BE^{2}$$CE^{2}$ + 2$AD^{2}$ – 2$DE^{2}$ $AB^{2}$$AC^{2}$$(BD+DE)^{2}$$(CD - DE)^{2}$ + 2$AD^{2}$ – 2$DE^{2}$

$BD^{2}+DE^{2}+2BD \times DE$ + $CD^{2}+DE^{2}+2CD \times DE$ + 2$AD^{2}$ – 2$DE^{2}$

$BD^{2}+2DE^{2}+2BD \times DE$ + $BD^{2}+2BD \times DE$ + 2$AD^{2}$ – 2$DE^{2}DE^{2}$

= 2$BD^{2}$ + 2$AD^{2}$

= 2($AD^{2}$ +$\frac{BC^{2}}{4}$)                (BC = 2BD)

### Proof

Therefore $AB^{2}+AC^{2}=2(AD^{2}+\frac{BC^{2}}{4})$                        (i)

similarly

$AB^{2}+BC^{2}=2(BE^{2}+\frac{AC^{2}}{4})$                            (ii)

$AC^{2}+BC^{2}=2(CF^{2}+\frac{AB^{2}}{4})$                               (iii)

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

$AB^{2}+AC^{2}+AB^{2}+BC^{2}+AC^{2}+BC^{2}$ = $2(AD^{2}+\frac{BC^{2}}{4})+2(BE^{2}+\frac{AC^{2}}{4}) +2(CF^{2}+\frac{AB^{2}}{4})$

2($AB^{2}+AC^{2}+ BC^{2}$) = 2($AD^{2}+\frac{BC^{2}}{4}+BE^{2}+\frac{AC^{2}}{4})+CF^{2}+\frac{AB^{2}}{4}$)

$AB^{2}+AC^{2}+ BC^{2}$ = $AD^{2}+\frac{BC^{2}}{4}+BE^{2}+\frac{AC^{2}}{4})+CF^{2}+\frac{AB^{2}}{4}$ $\frac{3}{4}$ ($AB^{2}+AC^{2}+ BC^{2}$) = $AD^{2}+BE^{2}+CF^{2}$

3 ($AB^{2}+AC^{2}+ BC^{2}$) = 4($AD^{2}+BE^{2}+CF^{2}$)