# Modern Geometry: Prove base angles of isosceles triangle are acute.

• Mar 20th 2009, 08:28 AM
ilikedmath
Modern Geometry: Prove base angles of isosceles triangle are acute.
Prove that the base angles of any isosceles triangle are acute.
Given this figure:

http://i43.tinypic.com/e7dqa8.jpg

Here is my work so far:
(So I need to show m
∠ACB < 90. And by showing that, I will also show that ∠B < 90 since ∠ACB http://i42.tinypic.com/1222p20.gif ∠B since they are the base angles of an isosceles triangle.)

Proof: Given isosceles triangle ABC.
Extend segment BC to ray BD by construction.
m∠ACB + m∠DCA = 180 by supplementary angle defn.
Assume
m∠ACB ≥ 90. ∠ACB http://i42.tinypic.com/1222p20.gif ∠ABC by base angles of isosceles triangle are congruent.
Then
m∠ACB + m∠ABC ≥ 180, but then the angle measure of triangle ABC will be > 180 which is a C! (contradiction) since a triangle's total angle sum is 180.
Therefore,
m∠ACB < 90 and since ∠ACB http://i42.tinypic.com/1222p20.gif ∠ABC, then also, m∠ABC < 90.
Therefore, the base angles of any isosceles triangle are acute. QED.

Is that an okay proof?(Wait)

Thank you for your time and help/suggestions/corrections.(Hi)
• Mar 20th 2009, 08:36 AM
running-gag
Hi

It's OK (Clapping)
• Mar 20th 2009, 09:21 AM
ilikedmath
Quote:

Originally Posted by running-gag
Hi

It's OK (Clapping)

Thank you!!!