Base Angle Theorem

The Base Angle Theorem states that in an isosceles triangle, the angles opposite the congruent sides are congruent.


Since the triangle only has three sides, the two congruent sides must be adjacent. Let them meet at vertex $A$.

Now we draw altitude $AD$ to $BC$. From the Pythagorean Theorem, $BD=CD$, and thus $\triangle ABD$ is congruent to $\triangle ACD$, and $\angle DBA=\angle DCA$. [asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H; A=(0,10); B=(-5,0); C=(5,0); D=(0,0); E=(1,1); F=(-1,1); G=(-1,0); H=(1,0); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(E--F); draw(E--H); draw(F--G); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S);[/asy]

Simpler Proof

We know that $\overline{AB} \cong \overline{AC}$ (given). By the reflexive property, we know that $\overline{BC} \cong \overline{CB}$. We know that $\overline{CA} \cong \overline{BA}$ (given). By SSS, we conclude that $\Delta ABC \cong \Delta ACB$. By CPCTC, we conclude that $\angle ABC \cong \angle ACB$.

[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H; A=(0,15); B=(-5,0); C=(5,0); draw(A--B); draw(B--C); draw(C--A); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); [/asy]

Even Simpler Proof

By the Law of Sines, we have $\tfrac{b}{\sin(B)}=\tfrac{c}{\sin(C)}$. We know $b=c$, so $\sin(B)=\sin(C)$. Then either $B=C$ or $B=180-C$, but the second case would imply $A=0^{\circ}$, so $B=C$.

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