# converse of isosceles triangle theorem

###### Theorem 1 ().

If $\triangle ABC$ is a triangle  with $D\in\overline{BC}$ such that any two of the following three statements are true:

1. 1.

$\overline{AD}$ is a median

2. 2.

$\overline{AD}$

3. 3.

$\overline{AD}$$\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 $\overrightarrow{CD}$. 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. ∎

Title converse of isosceles triangle theorem ConverseOfIsoscelesTriangleTheorem 2013-03-22 17:12:20 2013-03-22 17:12:20 Wkbj79 (1863) Wkbj79 (1863) 7 Wkbj79 (1863) Theorem msc 51-00 msc 51M04 IsoscelesTriangleTheorem AngleBisectorAsLocus