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 $B=C$ .

Page

Toolbox

Search

Difference between revisions of "Base Angle Theorem"

Latest revision as of 12:07, 20 February 2024

Proof

Simpler Proof

Even Simpler Proof

@@ Line 4: / Line 4: @@
 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 [[height]] <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>
+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;
@@ Line 27: / Line 27: @@
 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]]