In triangle ABC, prove that angle A is a right angle if and only if the length of the median from A to BC is exactly half the length of side BC.

Question

I drew it to scale because I was trying to notice something, but I cannot figure anything out.

Answer 1

"If": In $\triangle ABC$, let $E$ be the mid-point of $BC$, then $BE = AE = EC$. Then $A,B,C$ is on a circle with centre $E$.

And since $BEC$ is a straight line, it is also a diameter. $\angle A = 90^\circ$ because it is an angle in semi-circle.

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"Only if": Duplicate $\triangle A'CB \cong\triangle ABC$, with $A'$ and $A$ on different sides of line $BC$. Then $ABA'C$ is a rectangle because $\angle A = 90^\circ$.

The median from $A$ to the mid-point of $BC$ is half of the diagonal $AA'$ of the rectangle. In a rectangle, diagonals are of equal length and bisect each other.

Answer 2

An alternative. By the law of cosines, $$ 4m_a^2 = 2b^2+2c^2-a^2,\tag{1}$$ hence $4m_a^2=a^2$ implies: $$ b^2+c^2=a^2\tag{2}$$ from which $\widehat{BAC}=\frac{\pi}{2}$.