Modern Geometry: Prove base angles of isosceles triangle are acute.
Prove that the base angles of any isosceles triangle are acute.
Given this figure:
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)