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Difference between revisions of "Base Angle Theorem"

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== Even Simpler Proof ==
 
== Even Simpler Proof ==
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By the [[Law of Sines]], we have <math>\tfrac{b}{\sin(B)}=\tfrac{c}{\sin(C)}</math>. We know <math>b=c</math>, so <math>\sin(B)=\sin(C)</math>. Then either <math>B=C</math> or <math>B=180-C</math>, but the second case would imply <math>A=0^{\circ}</math>, so <math>B=C</math>.
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By the [[Law of Sines]], [https://artofproblemsolving.com/wiki/index.php/TOTO_SLOT_:_SITUS_TOTO_SLOT_MAXWIN_TERBAIK_DAN_TERPERCAYA TOTO SLOT] we have <math>\tfrac{b}{\sin(B)}=\tfrac{c}{\sin(C)}</math>. We know <math>b=c</math>, so <math>\sin(B)=\sin(C)</math>. Then either <math>B=C</math> or <math>B=180-C</math>, but the second case would imply <math>A=0^{\circ}</math>, so <math>B=C</math>.
 
[[Category:Theorems]]
 
[[Category:Theorems]]
 
[[Category:Geometry]]
 
[[Category:Geometry]]

Revision as of 17:03, 19 February 2024

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, TOTO SLOT 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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