As shown in the figure, H is the vertical center of triangle ABC, and O is the outer center of triangle ABC. OG⊥BC.

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。