Difference between revisions of "Base Angle Theorem"
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Since the triangle only has three sides, the two congruent sides must be adjacent. Let them meet at vertex <math>A</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 [[height]] <math>AD</math> to <math>BC</math>. From the [[Pythagorean Theorem]], <math>BD=CD</math>, and thus <math>\triangle ABD</math> is | + | 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> |
unitsize(5); defaultpen(fontsize(10)); | unitsize(5); defaultpen(fontsize(10)); | ||
pair A,B,C,D,E,F,G,H; | pair A,B,C,D,E,F,G,H; | ||
Revision as of 23:51, 13 February 2009
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 congruent 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)
