determining from angles that a triangle is isosceles
The following theorem holds in any geometry^{} in which ASA is valid. Specifically, it holds in both Euclidean geometry and hyperbolic geometry (and therefore in neutral geometry) as well as in spherical geometry.
Proof.
Let triangle $\mathrm{\u25b3}ABC$ have angles $\mathrm{\angle}B$ and $\mathrm{\angle}C$ congruent.
Since we have

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$\mathrm{\angle}B\cong \mathrm{\angle}C$

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$\overline{BC}\cong \overline{CB}$ by the reflexive property (http://planetmath.org/Reflexive^{}) of $\cong $ (note that $\overline{BC}$ and $\overline{CB}$ denote the same line segment^{})

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$\mathrm{\angle}C\cong \mathrm{\angle}B$ by the symmetric^{} property (http://planetmath.org/Symmetric) of $\cong $
we can use ASA to conclude that $\mathrm{\u25b3}ABC\cong \mathrm{\u25b3}ACB$. Since corresponding parts of congruent triangles are congruent, we have that $\overline{AB}\cong \overline{AC}$. It follows that $\mathrm{\u25b3}ABC$ is isosceles. ∎
In geometries in which ASA and SAS are both valid, the converse theorem of this theorem is also true. This theorem is stated and proven in the entry angles of an isosceles triangle.
Title  determining from angles that a triangle is isosceles 

Canonical name  DeterminingFromAnglesThatATriangleIsIsosceles 
Date of creation  20130322 17:12:23 
Last modified on  20130322 17:12:23 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  7 
Author  Wkbj79 (1863) 
Entry type  Theorem 
Classification  msc 51M04 
Classification  msc 5100 
Related topic  AnglesOfAnIsoscelesTriangle 