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[parent] isosceles triangle theorem (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 (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$

\begin{pspicture}(-3,-2)(3,3) \pspolygon(-2,-2)(0,2)(2,-2) \psline(-1.2,0.1)(-0.... ...put[r](-2.2,-2){$B$} \rput[a](0,-2.3){$D$} \rput[l](2.2,-2){$C$} \end{pspicture}
Proof. $ 1 \Rightarrow 2$: Since $ \overline{AD}$ is a median, $ \overline{BD} \cong \overline{CD}$. Since we have

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

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. $ \qedsymbol$

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.



"isosceles triangle theorem" is owned by Wkbj79.
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See Also: converse of isosceles triangle theorem


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Cross-references: perpendicular bisector, angles of an isosceles triangle, right angles, perpendicular, angles, congruent, CPCTC, angle bisector, altitude, median, the following are equivalent, neutral geometry, hyperbolic geometry, Euclidean geometry, SAS, AAS, SSS, isosceles triangle, geometries
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This is version 4 of isosceles triangle theorem, born on 2007-06-04, modified 2007-06-06.
Object id is 9523, canonical name is IsoscelesTriangleTheorem.
Accessed 5633 times total.

Classification:
AMS MSC: 51-00 (Geometry :: General reference works )
 51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries)
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