Prove that the medians of an equilateral tria | vedclass

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(A) Let $	riangle ABC$ be an equilateral triangle where $AB = BC = CA = a$. Let $AD$,$BE$,and $CF$ be the medians drawn to sides $BC$,$AC$,and $AB$ r

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(A) Let $	riangle ABC$ be an equilateral triangle where $AB = BC = CA = a$. Let $AD$,$BE$,and $CF$ be the medians drawn to sides $BC$,$AC$,and $AB$ respectively. In an equilateral triangle,the median is also the altitude. In $	riangle ABD$ and $	riangle BCE$: $AB = BC = a$ (sides of equilateral triangle). $\angle B = \angle B$ (common angle). $BD = BE$ is not the correct approach; let us use the Pythagorean theorem. In $	riangle ABD$,$\angle ADB = 90^\circ$. By Pythagoras theorem,$AD^2 = AB^2 - BD^2$. Since $D$ is the midpoint of $BC$,$BD = a/2$. $AD^2 = a^2 - (a/2)^2 = a^2 - a^2/4 = 3a^2/4$. So,$AD = (a\sqrt{3})/2$. Similarly,for median $BE$,in $	riangle BCE$,$BE^2 = BC^2 - CE^2 = a^2 - (a/2)^2 = 3a^2/4$. So,$BE = (a\sqrt{3})/2$. Similarly,$CF = (a\sqrt{3})/2$. Since $AD = BE = CF = (a\sqrt{3})/2$,the medians of an equilateral triangle are equal.

Prove that the medians of an equilateral triangle are equal.

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