Circle O is the inscribed circle of RT triangle ABC, DEF is the tangent point, and the extension line of de intersects the extension line of AC at g to prove BD = CG

Circle O is the inscribed circle of RT triangle ABC, DEF is the tangent point, and the extension line of de intersects the extension line of AC at g to prove BD = CG

In the triangle ABC connecting OD, OE, OB of ocrt, BD = be, OE = CF = Ce (because OB and OC are angular bisectors, a property of angular bisectors determines the congruence of triangles, such as OBD is equal to OBE). Let ob intersection de and H prove that beh is similar to OEB

Triangle ABC is an equilateral triangle. If angle abd = angle BCF = angle CAD, then triangle DEF is an equilateral triangle. Why?

Angle abd = angle BCF
Angle abd + angle DBC = 60 degrees
So angle FCD + angle DBC = 60 degrees
Because angle FCD + angle DBC = angle DFE [external angle theorem]
So the angle DFE = 60 degrees
Similarly, FDE = 60 degrees, def = 60 degrees
DEF is an equilateral triangle