You are here
Homeisosceles triangle theorem
Primary tabs
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 (Isosceles Triangle Theorem).
Let $\triangle ABC$ be an isosceles triangle such that $\overline{AB}\cong\overline{AC}$. Let $D\in\overline{BC}$. 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 $\angle BAC$
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 of $\cong$
we can use SSS to conclude that $\triangle ABD\cong\triangle ACD$. By CPCTC, $\angle ADB\cong\angle ADC$. Thus, $\angle ADB$ and $\angle ADC$ are supplementary 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, $\angle ADB$ and $\angle ADC$ are right angles and therefore congruent. Since we have

$\angle B\cong\angle C$ by the theorem on angles of an isosceles triangle

$\angle ADB\cong\angle ADC$
we can use AAS to conclude that $\triangle ABD\cong\triangle ACD$. By CPCTC, $\angle BAD\cong\angle CAD$. It follows that $\overline{AC}$ is the angle bisector of $\angle BAC$.
$3\Rightarrow 1$: Since $\overline{AD}$ is an angle bisector, $\angle BAD\cong\angle CAD$. Since we have

$\overline{AB}\cong\overline{AC}$

$\angle BAD\cong\angle CAD$

$\overline{AD}\cong\overline{AD}$ by the reflexive property of $\cong$
we can use SAS to conclude that $\triangle ABD\cong\triangle ACD$. By CPCTC, $\overline{BD}\cong\overline{CD}$. It follows that $\overline{AD}$ is a median. ∎
Remark: Another equivalent 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.
Mathematics Subject Classification
5100 no label found51M04 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
Recent Activity
new question: Prove that for any sets A, B, and C, An(BUC)=(AnB)U(AnC) by St_Louis
Apr 20
new image: informationtheoreticdistributedmeasurementdds.png by rspuzio
new image: informationtheoreticdistributedmeasurement4.2 by rspuzio
new image: informationtheoreticdistributedmeasurement4.1 by rspuzio
new image: informationtheoreticdistributedmeasurement3.2 by rspuzio
new image: informationtheoreticdistributedmeasurement3.1 by rspuzio
new image: informationtheoreticdistributedmeasurement2.1 by rspuzio
Apr 19
new collection: On the InformationTheoretic Structure of Distributed Measurements by rspuzio
Apr 15
new question: Prove a formula is part of the Gentzen System by LadyAnne
Mar 30
new question: A problem about Euler's totient function by mbhatia