Browse Questions

Using vector method, prove that if two medians of a triangle are equal, then it is an isosceles.

Toolbox:
• Median is the segment joining mid point of any side with its opposite vertex.
• Section formula: If D is mid point of BC then, $\overrightarrow {OD}=\frac{\overrightarrow {OB}+\overrightarrow {OC}}{2}$
• $\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}$
• A triangle is isosceles if two sides are equal.
• $|\overrightarrow a+\overrightarrow b|^2=|\overrightarrow a|^2+|\overrightarrow b|^2+2\overrightarrow a.\overrightarrow b$
• $|\overrightarrow a-\overrightarrow b|^2=|\overrightarrow a|^2+|\overrightarrow b|^2-2\overrightarrow a.\overrightarrow b$
Let ABC be triangle where
D is mid point of BC, E is mid point of AC and F is mid point of AB
Let $\overrightarrow {OA}=\overrightarrow a,\:\:\overrightarrow {OB}=\overrightarrow b\:\:and\:\:\overrightarrow {OC}=\overrightarrow c$
be the position vectors of the points A,B and C respectively.
$\Rightarrow$ the sides AB,BC and AC are given by
$\overrightarrow {AB}=\overrightarrow b-\overrightarrow a,\:\:\overrightarrow {BC}=\overrightarrow c-\overrightarrow b\:and\:\overrightarrow {AC}=\overrightarrow c-\overrightarrow a$
From section formula we can say that $\overrightarrow {OD}=\frac{\overrightarrow b+\overrightarrow c}{2}$,
$\overrightarrow {OE}=\frac{\overrightarrow a+\overrightarrow c}{2}$ and $\overrightarrow {OF}=\frac{\overrightarrow a+\overrightarrow b}{2}$
AD, BE and CF are the medians.
According to the question it is given that two medians are equal.
Let the two equal medians be $\overrightarrow {AD}$ and $\overrightarrow {BE}$
$\Rightarrow |\overrightarrow {AD}|=|\overrightarrow {BE}|$
$\Rightarrow |\overrightarrow {AD}|^2=|\overrightarrow {BE}|^2$
$\Rightarrow\:|\overrightarrow {OD}-\overrightarrow {OA}|^2=|\overrightarrow {OE}-\overrightarrow {OB}|^2$
$\Rightarrow |\frac{\overrightarrow b+\overrightarrow c}{2}-\overrightarrow a|^2=|\frac{\overrightarrow a+\overrightarrow c}{2}-\overrightarrow b|^2$
$\Rightarrow\:|\frac{\overrightarrow b+\overrightarrow c}{2}|^2+|\overrightarrow a|^2-2(\frac{\overrightarrow b+\overrightarrow c}{2}).\overrightarrow a$
$=|\frac{\overrightarrow a+\overrightarrow c}{2}|^2+|\overrightarrow b|^2-2(\frac{\overrightarrow a+\overrightarrow c}{2}).\overrightarrow b$
$\Rightarrow\:\frac{|\overrightarrow b|^2}{4}+\frac{|\overrightarrow c|^2}{4}+\frac{\overrightarrow b.\overrightarrow c}{2}+|\overrightarrow a|^2-\overrightarrow b.\overrightarrow a-\overrightarrow c.\overrightarrow a$
$=\frac{|\overrightarrow a|^2}{4}+\frac{|\overrightarrow c|^2}{4}+\frac{\overrightarrow a.\overrightarrow c}{2}+|\overrightarrow b|^2-\overrightarrow a.\overrightarrow b-\overrightarrow c.\overrightarrow b$
$\Rightarrow \frac{3}{4}|\overrightarrow a|^2+\frac{3}{2}\overrightarrow b.\overrightarrow c=\frac{3}{4}|\overrightarrow b|^2+\frac{3}{2}\overrightarrow a.\overrightarrow c$
$\Rightarrow |\overrightarrow a|^2-2\overrightarrow a.\overrightarrow c=|\overrightarrow b|^2-2\overrightarrow b.\overrightarrow c$
Adding $|\overrightarrow c|^2$ on both sides we get
$\Rightarrow |\overrightarrow a|^2-2\overrightarrow a.\overrightarrow c+|\overrightarrow c|^2=|\overrightarrow b|^2-2\overrightarrow b.\overrightarrow c+|\overrightarrow c|^2$
$\Rightarrow |\overrightarrow a-\overrightarrow c|^2=|\overrightarrow b-\overrightarrow c|^2$
$\Rightarrow \:|\overrightarrow {CA}|=|\overrightarrow {CB}|$
$\Rightarrow$ the triangle is isosceles.
edited Apr 28, 2013