converse of isosceles triangle theorem

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

Theorem 1 ().

If $\mathrm{△}\mathit{}A\mathit{}B\mathit{}C$ is a triangle with $D\mathrm{\in }\overline{B\mathit{}C}$ such that any two of the following three statements are true:

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

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

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

then $\mathrm{△}\mathit{}A\mathit{}B\mathit{}C$ is isosceles.

Proof.

First, assume 1 and 2 are true. Since $\overline{AD}$ is a median, $\overline{BD}\cong \overline{CD}$. 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

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

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

• $\overline{BD}\cong \overline{CD}$

we can use SAS to conclude that $\mathrm{△}ABD\cong \mathrm{△}ACD$. By CPCTC, $\overline{AB}\cong \overline{AC}$.

Next, assume 2 and 3 are true. 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 $\overline{AD}$ is an angle bisector, $\mathrm{\angle }BAD\cong \mathrm{\angle }CAD$. Since we have

• $\mathrm{\angle }ADB\cong ADC$

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

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

we can use ASA to conclude that $\mathrm{△}ABD\cong \mathrm{△}ACD$. By CPCTC, $\overline{AB}\cong \overline{AC}$.

Finally, assume 1 and 3 are true. Since $\overline{AD}$ is an angle bisector, $\mathrm{\angle }BAD\cong \mathrm{\angle }CAD$. Drop perpendiculars from $D$ to the rays $\overrightarrow{AB}$ and $\overrightarrow{CD}$. the intersections as $E$ and $F$, respectively. Since the length of $\overline{DE}$ is at most $\overline{BD}$, we have that $E\in \overline{AB}$. (Note that $E\ne A$ and $E\ne B$ are not assumed.) Similarly $F\in \overline{AC}$.

Since we have

• $\mathrm{\angle }AED\cong \mathrm{\angle }AFD$

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

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

we can use AAS to conclude that $\mathrm{△}ADE\cong \mathrm{△}ADF$. By CPCTC, $\overline{DE}\cong \overline{DF}$ and $\mathrm{\angle }ADE\cong \mathrm{\angle }ADF$.

Since $\overline{AD}$ is a median, $\overline{BD}\cong \overline{CD}$. Recall that SSA holds when the angles are right angles. Since we have

• $\overline{BD}\cong \overline{CD}$

• $\overline{DE}\cong \overline{DF}$

• $\mathrm{\angle }BED$ and $\mathrm{\angle }CFD$ are right angles

we can use SSA to conclude that $\mathrm{△}BDE\cong \mathrm{△}CDF$. By CPCTC, $\mathrm{\angle }BDE\cong \mathrm{\angle }CDF$.

Recall that $\mathrm{\angle }ADE\cong \mathrm{\angle }ADF$ and $\mathrm{\angle }BDE\cong \mathrm{\angle }CDF$. Thus, $\mathrm{\angle }ADB\cong \mathrm{\angle }ADC$. Since we have

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

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

• $\overline{BD}\cong \overline{CD}$

we can use SAS to conclude that $\mathrm{△}ABD\cong \mathrm{△}ACD$. By CPCTC, $\overline{AB}\cong \overline{AC}$.

In any case, $\overline{AB}\cong \overline{AC}$. It follows that $\mathrm{△}ABC$ is isosceles.∎