Difference between revisions of "Base Angle Theorem"

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The '''Hinge Theorem''' states that in an [[isosceles triangle]], the measures of the angles opposite the equal-measuring sides are equal.
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The '''Base Angle Theorem''' states that in an [[isosceles triangle]], the angles opposite the congruent sides are congruent.
  
==Proof==
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== Proofs ==
Since the triangle only has three sides, the two equal-measuring sides must be adjacent. Let them meet at vertex <math>A</math>.
 
  
Now we draw [[height]] <math>AD</math> to <math>BC</math>. From the [[Pythagorean Theorem]], <math>BD=CD</math>, and thus <math>\triangle ABD</math> is similar to <math>\triangle ACD</math>, and <math>\angle DBA=\angle DCA</math>. <asy>
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== Proof 1 ==
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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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== Proof 2 ==
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We know that <math>\overline{AB} \cong \overline{AC}</math> (given). By the reflexive property, we know that <math>\overline{BC} \cong \overline{CB}</math>. We know that <math>\overline{CA} \cong \overline{BA}</math> (given).  By SSS, we conclude that <math>\Delta ABC \cong \Delta ACB</math>. By CPCTC, we conclude that <math>\angle ABC \cong \angle ACB</math>.
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== Proof 3 ==
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Since the triangle only has three sides, the two congruent sides must be adjacent. Let them meet at vertex <math>A</math>.
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Now we draw [[altitude]] <math>AD</math> to <math>BC</math>. From the [[Pythagorean Theorem]], <math>BD=CD</math>, and thus <math>\triangle ABD</math> is congruent to <math>\triangle ACD</math>, and <math>\angle DBA=\angle DCA</math>. <asy>
 
unitsize(5); defaultpen(fontsize(10));
 
unitsize(5); defaultpen(fontsize(10));
 
pair A,B,C,D,E,F,G,H;
 
pair A,B,C,D,E,F,G,H;
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[[Category:Theorems]]
 
[[Category:Theorems]]
 
[[Category:Geometry]]
 
[[Category:Geometry]]
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{{stub}}

Latest revision as of 17:41, 17 January 2025

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

Proofs

Proof 1

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 $B=C$.

Proof 2

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 $\angle ABC \cong \angle ACB$.

Proof 3

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] This article is a stub. Help us out by expanding it.