Let ABC be an equilateral triangle with point D on AC.
Create a new triangle with side lengths AD, DC, BD.
Prove one angle of this new triangle is 120 degrees.
(In reply to
Solution by Brian Smith)
Same setup as before: let a=AD, b=BD, and c=CD. Then AB=BC=a+c.
But this time I realized I could take the law of cosines on one of the known 60 degree angles. On just triangle ABD:
b^2 = (a+c)^2+a^2-2a(a+c)*cos(60)
b^2 = a^2+2ac+c^2-a^2-ac
b^2 = a^2 + c^2 + ac
b^2 = a^2 + c^2 - 2ac*cos(120)
Much more quickly we arrive at the same law of cosines on a 120 degree angle triangle with sides of length a, b, and c with the 120 degree angle opposite of b.
re: Solution - Quicker
