Bisector, median and altitude

The symmetry of the isosceles triangle causes three notable lines, which in a generic triangle are distinct, to overlap here into a single one.

Theorem — Bisector, median and altitude in the isosceles triangle

In an isosceles triangle $ABC$ with $AC\cong BC$, the bisector of the apex angle $\widehat{C}$ is also the median (it passes through the midpoint of $AB$) and the altitude (it is perpendicular to $AB$).

In the isosceles triangle the bisector from $C$ is simultaneously the median and the altitude relative to the base $AB$.

Proof

We show that the bisector from $C$ falls at the midpoint of $AB$ and is perpendicular to it.

1. Construction. Let $K$ be the foot of the bisector of $\widehat{C}$ on $AB$.
2. I consider. In the triangles $AKC$ and $BKC$ we find: $CK$ in common, $\widehat{ACK}\cong\widehat{BCK}$ (bisector), $AC\cong BC$ (hypothesis).
3. I deduce. By the first criterion: $\triangle AKC\cong\triangle BKC$.
4. Verified. From the congruence it follows that $AK\cong BK$: $K$ is the midpoint of $AB$, hence $CK$ is the median.
5. I deduce. Moreover $\widehat{AKC}\cong\widehat{BKC}$. Since they are supplementary and congruent, each equals $\dfrac{\pi}{2}$: $CK$ is the altitude.

In short, in the isosceles triangle the bisector, median and altitude from the apex coincide. $\blacksquare$

Links

Topics: Euclidean geometry
Concepts: Altitude · Bisector · Congruence criteria · Proof · Median · Line · Isosceles triangle
Skills: Proving · Synthetic geometry