# Base Angle Theorem

(Redirected from Hinge theorem)

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

## Proof

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 $B=C$.