Difference between revisions of "Base Angle Theorem"
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| − | The '''Base Angle Theorem''' states that in an [[isosceles triangle]], | + | 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 [[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> | 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> | ||
Revision as of 21:44, 20 September 2008
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 height 
to 
. From the Pythagorean Theorem, 
, and thus 
is similar to 
, and 
. ![[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]](http://latex.artofproblemsolving.com/7/f/6/7f66cb51375e207e519d2fb8340d23232cff7953.png)
