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

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 \triangleBDF and \triangleCDG

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

DG = DF         ( by construction )

\angleBDF = \angleCDG              ( vertically opposite angle )

so \triangleBDF \cong \triangleCDG

therefore \angleBFD = \angleCGD

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

Now in \triangleABF

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

 

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

Proof

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

Area of \triangleABD = \frac{1}{2}\times base\times height

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

Area of \triangleADC = \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 \triangleABD = Area of \triangleADC

 

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

Proof

In \triangleBOC, OD is the median of triangle

\Rightarrow area of \triangleBOD = area of \triangleCOD = y

In \triangleAOC, OE is the median of triangle

\Rightarrow area of \triangleAOE = area of \triangleCOE = x

In \triangleAOB, OF is the median of triangle

\Rightarrow area of \triangleAOF = area of \triangleBOF = z

In \triangleABC, AD is the median of triangle

\Rightarrow area of \triangleABD = area of \triangleACD

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

\Rightarrow 2z = 2x

\Rightarrow z = x                           (i)

In \triangleABC, BE is the median of triangle

\Rightarrow area of \triangleABE = area of \triangleCBE

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

\Rightarrow z = y                        (ii)

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

Proof

In \triangleABD, \angleABD = 90^{\circ} AB^{2} + BD^{2} = AD^{2}                         (1)

Similarly, in \triangleBCE, \angleEBC = 90^{\circ} BC^{2} + BE^{2} = CE^{2}                             (2)

Adding equ. (1) and (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} ) = 5AC^{2}

 

5. If ABC is a triangle in which AD is the median on the side BC,  then AB^{2}+AC^{2}=2(AD^{2}+\frac{BC^{2}}{4})

Proof 

Draw AE\perpBC

In \triangleABE, by pythagorus theorem

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

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

Adding equation (i) and (ii)

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

Now in \triangleADE

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

Put this value in equation (iii)

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

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

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

= 2BD^{2} + 2AD^{2}

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

 

6. If in a \triangleABC, AD, BE and CF are the medians then 4 (AD^{2}+BE^{2}+CF^{2}) = 3(AB^{2}+BC^{2}+CA^{2})

Proof 

Since AD is the median

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})

 

 

 

 

 

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