# Median of a Triangle

When we were in high school we never thought that there is actually a formula to solve for the median of a triangle. Probably because the derivation of this formula is by cosine law and geometry is being teach in 3rd year (grade 9).

Well it’s time to move on. I’m done with my high school life. This topic may guide high schoolers whenever they need it for school or contest.

Median of a triangle is a line connecting from one vertex to the midpoint of the side opposite to it.

median of a triangle

In the figure above, $d$ is the median of $\triangle ABC$

Derivation: Using the same figure above

Using cosine law in $\triangle ABC$ we have the following

$b^2=a^2+c^2-2ac\cos B$  (1)

$c^2=a^2+b^2-2ab\cos C$ (2)

Addig (1) and (2) we have,

$b^2+ c^2=a^2+c^2-2ac\cos B+a^2+b^2-2ab\cos C$

$0=2a^2-2ac\cos B-2ab\cos C$

$2a^2=2ac\cos B+2ab\cos C$

$a=c\cos B+b\cos C$  (3)

Using cosine law in $\triangle ABD$,

$d^2=(\displaystyle\frac{a}{2})^2+c^2-2\displaystyle\frac{a}{2}c\cos B$

$d^2=(\displaystyle\frac{a}{2})^2+c^2-ac\cos B$

$\cos B=\displaystyle\frac{c^2+\frac{a^2}{4}-d^2}{ac}$  (4)

Using cosine law in $\triangle ACD$,

$d^2=(\displaystyle\frac{a}{2})^2+b^2-2\displaystyle\frac{a}{2}b\cos C$

$d^2=(\displaystyle\frac{a}{2})^2+b^2-ab\cos C$

$\cos C=\displaystyle\frac{b^2+\frac{a^2}{4}-d^2}{ab}$  (5)

Substitute both (4) and (5) to (3)

$a=c\cos B+b\cos C$

$a=c\displaystyle(\displaystyle\frac{c^2+\frac{a^2}{4}-d^2}{ac})+b\displaystyle(\displaystyle\frac{b^2+\frac{a^2}{4}-d^2}{ab})$

$a=\displaystyle(\displaystyle\frac{c^2+\frac{a^2}{4}-d^2}{a})+\displaystyle(\displaystyle\frac{b^2+\frac{a^2}{4}-d^2}{a})$

$a^2=b^2+c^2+\displaystyle\frac{a^2}{2}-2d^2$

Solving for median (d) we have

$2d^2= b^2+c^2+\displaystyle\frac{a^2}{2}-a^2$

Simplifying this further we have

$d_a=\displaystyle\frac{\sqrt{2b^2+2c^2-a^2}}{2}$

Generally, the length of the median from one vertex to the opposite side is half the square root of the difference of the sum of twice the sum of squares of adjacent sides and the opposite side. Yes! That’s confusing. The latter formula can be interchanged in the following manner.

$d_b=\displaystyle\frac{\sqrt{2a^2+2c^2-b^2}}{2}$

$d_c=\displaystyle\frac{\sqrt{2b^2+2a^2-c^2}}{2}$

Proceed to next page for worked problems

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