isosceles triangle theorem

The following theorem holds in geometries in which isosceles triangle can be defined and in which SSS, AAS, and SAS are all valid. Specifically, it holds in Euclidean geometry and hyperbolic geometry (and therefore in neutral geometry).

Theorem 1 ().

Let $\mathrm{△}\mathit{}A\mathit{}B\mathit{}C$ be an isosceles triangle such that $\overline{A\mathit{}B}\mathrm{\cong }\overline{A\mathit{}C}$. Let $D\mathrm{\in }\overline{B\mathit{}C}$. Then the following are equivalent:

1. $\overline{AD}$ is a median

2. $\overline{AD}$ is an altitude

3. $\overline{AD}$ is the angle bisector of $\mathrm{\angle }BAC$

Proof.

$1\Rightarrow 2$: Since $\overline{AD}$ is a median, $\overline{BD}\cong \overline{CD}$. Since we have

• $\overline{AB}\cong \overline{AC}$

• $\overline{BD}\cong \overline{CD}$

• $\overline{AD}\cong \overline{AD}$ by the reflexive property (http://planetmath.org/Reflexive) of $\cong$

we can use SSS to conclude that $\mathrm{△}ABD\cong \mathrm{△}ACD$. By CPCTC, $\mathrm{\angle }ADB\cong \mathrm{\angle }ADC$. Thus, $\mathrm{\angle }ADB$ and $\mathrm{\angle }ADC$ are supplementary (http://planetmath.org/SupplementaryAngle) congruent angles. Hence, $\overline{AD}$ and $\overline{BC}$ are perpendicular. It follows that $\overline{AD}$ is an altitude.

$2\Rightarrow 3$: Since $\overline{AD}$ is an altitude, $\overline{AD}$ and $\overline{BC}$ are perpendicular. Thus, $\mathrm{\angle }ADB$ and $\mathrm{\angle }ADC$ are right angles and therefore congruent. Since we have

• $\mathrm{\angle }B\cong \mathrm{\angle }C$ by the theorem on angles of an isosceles triangle

• $\mathrm{\angle }ADB\cong \mathrm{\angle }ADC$

• $\overline{AD}\cong \overline{AD}$ by the reflexive property of $\cong$

we can use AAS to conclude that $\mathrm{△}ABD\cong \mathrm{△}ACD$. By CPCTC, $\mathrm{\angle }BAD\cong \mathrm{\angle }CAD$. It follows that $\overline{AC}$ is the angle bisector of $\mathrm{\angle }BAC$.

$3\Rightarrow 1$: Since $\overline{AD}$ is an angle bisector, $\mathrm{\angle }BAD\cong \mathrm{\angle }CAD$. Since we have

• $\overline{AB}\cong \overline{AC}$

• $\mathrm{\angle }BAD\cong \mathrm{\angle }CAD$

• $\overline{AD}\cong \overline{AD}$ by the reflexive property of $\cong$

we can use SAS to conclude that $\mathrm{△}ABD\cong \mathrm{△}ACD$. By CPCTC, $\overline{BD}\cong \overline{CD}$. It follows that $\overline{AD}$ is a median.∎

Remark: Another equivalent (http://planetmath.org/Equivalent3) condition for $\overline{AD}$ is that it is the perpendicular bisector of $\overline{BC}$; however, this fact is usually not included in the statement of the Isosceles Triangle Theorem.