Difference between revisions of "Base Angle Theorem"
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− | The '' | + | The '''Base Angle Theorem''' states that in an [[isosceles triangle]], the angles opposite the congruent sides are congruent. |
==Proof== | ==Proof== | ||
− | Since the triangle only has three sides, the two | + | Since the triangle only has three sides, the two congruent sides must be adjacent. Let them meet at vertex <math>A</math>. |
+ | 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)); | ||
+ | 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 <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>. | ||
+ | |||
+ | <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 <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]] |
Latest revision as of 13:07, 20 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 .
Now we draw altitude to
. From the Pythagorean Theorem,
, and thus
is congruent to
, and
.
Simpler Proof
We know that (given). By the reflexive property, we know that
. We know that
(given). By SSS, we conclude that
. By CPCTC, we conclude that
.
Even Simpler Proof
By the Law of Sines, we have . We know
, so
. Then either
or
, but the second case would imply
, so
.