ABC is a right angled triangle, with the right angle at C. Equilateral triangles BCD, CAE and ABF are constructed on the outside of triangle ABC.
Prove that AD = BE
Let A be at (0,a); B at (b,0), and C at (0,0).
Then D is at (b/2 , - sqrt(3)*b/2)
And E is at (- sqrt(3)*a/2 , a/2)
Distance(AD)^2 = (1/4)b^2 + a^2 + (3/4)b^2 + sqrt(3)*ab/2
So Distance(BE)^2 = b^2 + (3/4)a^2 + sqrt(3)*ab/2 + (1/4)a^2
So both distances are a^2 + b^2 + sqrt(3)*ab/2
Posted by Larry
on 2019-04-23 09:47:39