Verification: AH=2GO
It is proved that the circumscribed circle O connecting BO and extending the intersecting triangle ABC at point F connects AF and CF, then:
FA⊥AB,FC⊥BC.
(The circumferential angle of the diameter is a right angle)
What comes from H is experience: AD⊥BC,CE⊥AB.
So: AF‖EC, AD‖FC, that is, the quadrilateral AHCF is a parallelogram. Yes, AH=FC
Because: OG⊥BC,FC⊥BC
So: OG‖FC
And o is the midpoint of BF, so there is OG=( 1/2)FC.
Therefore: OG=( 1/2)AH,
that is
AH=2GO。