Difference between revisions of "Base Angle Theorem"

Latest revision as of 12:07, 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 $A$ .

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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Difference between revisions of "Base Angle Theorem"

Latest revision as of 12:07, 20 February 2024

Proof

Simpler Proof

Even Simpler Proof

@@ Line 1: / Line 1: @@
-The ''Hinge theorem''' states that in an [[isosceles triangle]], the measures of the angles opposite the equal-measuring sides are equal.
+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 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>.
+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:Geometry]]