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)