Draw ∇ABC
Extend BA to D and BC to E
Construct bisectors of ^GAC and ^ECA which intersect at F
Construct perpendicular from BD to F, meeting BD at G
Construct perpendicular from BE to F, meeting BE at H
Construct perpendicular from AC to F, meeting AC at J
Consider ∇AFG and ∇AFJ
^GAF = ^JAF (Construct)
^AGF = 90º = ^AJF (Construct)
∴ ^AFG = ^AFJ (Angles of triangle equal 2 rt. angles.)
AF = AF (common)
∴ ∇AGF = ∇AJF (2 angles and inc. side equal)
∴ GF = JF
Similarly considering ∇CJF and ∇CHF
JF = HF
∴ GF = JF = HF
Join B to F
Consider ∇BGF and ∇BHF
BF = BF (Common)
^BGF = 90º = ^BHF (Construct)
GF = HF (Proven above)
∴ ∇BGF = ∇BHF (Rt. ang. triangle, hypotenuse and 1 side equal)
∴ ^GBF = ^HBF