Difference between revisions of "Base Angle Theorem"

m (Proof)
Line 26: Line 26:
 
label("$C$",C,SE);
 
label("$C$",C,SE);
 
label("$D$",D,S);</asy>
 
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>
  
 
[[Category:Theorems]]
 
[[Category:Theorems]]
 
[[Category:Geometry]]
 
[[Category:Geometry]]

Revision as of 13:39, 17 January 2016

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]