This is outcome of boredom in managment classroom 🙂
Consider triangle ABC, right angled at B.
To prove AB^2+BC^2=AC^2
Draw circle X with AB as diameter. This circle cuts AC at O. Construct another circle Y with BC as diameter. Circle cuts AC at P.
From properties of Tangent and circle
Now AB is tangent to circle Y at B. Hence
AB^2 = (AP)(AC) ..(i)
Similarly, BC is tangent to circle X at B.
BC^2 = (OC)(AC) ..(ii)
Adding (i) and (ii)
AB^2 + BC^2 = (AP)(AC) + (OC)(AC) = (AC)(AP + OC)
Now since AB is diameter to Circle X, Angle AOB is right angled.
Similarly P is on circle Y with BC diameter. Hence angle BPC is right angled. Thus BP as well as BO is perpendicular to line AC; which is not possible unless they are same point. Hence OP = 0.
AP + OC = AC ..(iii)
Hence AB^2 + BC^2 = AC^2