# Why is the angle exactly 1/2 of one of the angles in the isosceles triangle

Why is the angle exactly 1/2 of one of the angles in the isosceles triangle
I'm sure there's a simple theorem that answers my question, but I'm unsure as to how to formulate the problem (not easy to describe geometry with words): It's a isosceles triangle where you draw a normal line on one of the "hypotenuse" sides, and then you make a right angled triangle out of that. My question is why is the 30 degree angle exactly 1/2 of the 60 degree angles we find in the isosceles triangle?
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getrdone07tl
Step 1
Because if you have a right triangle, with acute angles $\alpha ,\beta$, then $\left(\alpha +\beta \right)={90}^{\circ }$.
Step 2
When $\alpha ={60}^{\circ }=\frac{2}{3}×{90}^{\circ },$ then $\beta ={90}^{\circ }-{60}^{\circ }={90}^{\circ }\left(1-\frac{2}{3}\right)$
$={90}^{\circ }\left(\frac{1}{3}\right)=\left(\frac{1}{2}\right)×\alpha .$