Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles

Theorem
Let $\triangle ABC$ be a triangle.

Then:
 * the altitude from $AB$ to $C$
 * the median from $AB$ to $C$
 * the perpendicular bisector of $AB$
 * are all the same straight line


 * $\triangle ABC$ is isosceles where $AB$ is the base.
 * $\triangle ABC$ is isosceles where $AB$ is the base.

Necessary Condition
Let $\triangle ABC$ be an isosceles triangle whose base is $AB$.

Let $D$ be the midpoint of $\triangle ABC$.


 * IsoscelesAltitudeMedianPerpBis.png

By definition of isosceles triangle, $AC = BC$.

We have $AD = DB$ by construction, and $CD$ is common.

Sufficient Condition
Let $\triangle ABC$ be such that:
 * the altitude from $AB$ to $C$
 * the median from $AB$ to $C$
 * the perpendicular bisector of $AB$
 * are all the same straight line.

Suppose $\triangle ABC$ is not an isosceles triangle such that $AB$ is the base.


 * AltitudeMedianPerpendicularBisector.png