Draw ∇ABC with B a right angle.
Join B to hypotenuse mid point D
∴ AD = BD (construct)
Draw perpendicular ED from AB, and FD from BC
^ABC = 90º (Given)
= ^DFC (Construct)
∴ AB is parallel DF (Corr. angles equal)
Similarly ED is parallel DF
Consider ∇AED and ∇DFC
AD = DC (Construct)
^EAD = ^FDC (Corr. angles AB parallel DF)
^ADE = 180º-90º-^EAD (Angles in Triangle = 180º)
= 180º-90º-^FDC (Angles in Triangle = 180º)
= ^DCF
∴ ∇AED = ∇DFC (2 sides & incl. angle equal)
∴ ED = FC
EDFB is rectangle (Opp. sides parallel all angles 90º)
∴ ED = BF (Opp. sides of rectangle equal)
Consider ∇BDF and ∇CDF
BF = ED (Proven)
FC = ED (Proven)
∴ BF = FC
^BFD = 90 (Construct)
= ^CFD (Straight line = 180)
DF = DF (Common)
∴ ∇BFD = ∇CFD (2 angles & inc. angle equal)
∴ BD = CD
AD = DC (Given)
∴ AD = BD = CD