Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles

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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

if and only if:

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


Proof

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.

$\Box$


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


Aiming for a contradiction, suppose $D = E$.

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.

$\blacksquare$