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Theorem (Converse of Isosceles Triangle Theorem)   If $\triangle ABC$ is a triangle with $D \in \overline{BC}$ 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 $\angle BAC$

then $\triangle ABC$ 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, $\angle ADB$ and $\angle ADC$ are right angles and therefore congruent. Since we have

• $\overline{AD} \cong \overline{AD}$ by the reflexive property of $\cong$
• $\angle ADB \cong \angle ADC$
• $\overline{BD} \cong \overline{CD}$

we can use SAS to conclude that $\triangle ABD \cong \triangle ACD$ . 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, $\angle ADB$ and $\angle ADC$ are right angles and therefore congruent. Since $\overline{AD}$ is an angle bisector, $\angle BAD \cong \angle CAD$ . Since we have

• $\angle ADB \cong ADC$
• $\overline{AD} \cong \overline{AD}$ by the reflexive property of $\cong$
• $\angle BAD \cong \angle CAD$

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

Finally, assume 1 and 3 are true. Since $\overline{AD}$ is an angle bisector, $\angle BAD \cong \angle CAD$ . Drop perpendiculars from $D$ to the rays $\overrightarrow{AB}$ and $\overrightarrow{CD}$ . Label 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 $E \neq A$ and $E \neq B$ are not assumed.) Similarly $F \in \overline{AC}$ .

Since we have

• $\angle AED \cong \angle AFD$
• $\angle BAD \cong \angle CAD$
• $\overline{AD} \cong \overline{AD}$ by the reflexive property of $\cong$

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

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

• $\overline{BD} \cong \overline{CD}$
• $\overline{DE} \cong \overline{DF}$
• $\angle BED$ and $\angle CFD$ are right angles

we can use SSA to conclude that $\triangle BDE \cong \triangle CDF$ . By CPCTC, $\angle BDE \cong \angle CDF$ .

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

• $\overline{AD} \cong \overline{AD}$ by the reflexive property of $\cong$
• $\angle ADB \cong \angle ADC$
• $\overline{BD} \cong \overline{CD}$

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

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

converse of isosceles triangle theorem is owned by Warren Buck.