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 $AB$.


 * IsoscelesAltitudeMedianPerpBis.png

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

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

So by Triangle Side-Side-Side Equality:
 * $\triangle ACD = \triangle BCD$

From Two Angles on Straight Line make Two Right Angles:
 * $\angle ADC + \angle BDC$ equals two right angles

and as $\angle ADC = \angle BDC$ they are each both right angles.

Thus $CD$ is the perpendicular bisector of $AB$.

By definition:
 * As $D$ is the midpoint of $AB$, $CD$ is the median from $AB$ to $C$.
 * As $CD$ is perpendicular to $AB$ and passes through $C$, which is a vertex of $\triangle ABC$, $CD$ is the altitude from $AB$ to $C$.

Thus all three lines coincide.

Converse Statement
Let $AC \ne BC$ in $\triangle ABC$.

Let $CD$ be the altitude from $AB$ to $C$.

From Triangle with Two Equal Angles is Isosceles, it follows that $\angle CAB \ne \angle CBA$.

Thus $\triangle ACD \ne \triangle BCD$.

Let $E$ be the midpoint of $AB$.

Let $EF$ be the perpendicular bisector of $AB$.

By definition, $CE$ is the median from $AB$ to $C$.


 * AltitudeMedianPerpendicularBisector.png

$D = E$.

Then:
 * $AD = BD$
 * $AC = BC$
 * $CD$ is common.

So by Triangle Side-Side-Side Equality:
 * $\triangle ACD = \triangle BCD$

But we have that $\angle CAB \ne \angle CBA$.

So by Proof by Contradiction it follows that $D \ne E$.

It follows that $CD$, $CE$ and $EF$ are all different lines.