As shown in the figure, it is known that ad, be and CF are respectively the heights of the three sides of △ ABC, h is the perpendicular, and the extension line of ad intersects the circumscribed circle of △ ABC at point G

As shown in the figure, it is known that ad, be and CF are respectively the heights of the three sides of △ ABC, h is the perpendicular, and the extension line of ad intersects the circumscribed circle of △ ABC at point G

Connecting CG, ∵ ad ⊥ BC, ∵ ABC + ⊥ gab = 90 ° we can get ∵ ABC + ⊥ FCB = 90 ° so as to get ∵ gab = ⊥ FCB = 90 ° - ⊥ ABC, and ∵ gab and ⊥ GCB are the same opposite arc BG, ∵ gab = ⊥ GCB, we can get ∵ GCB = ⊥ FCB, ∵ CD ⊥ GH, that is, CD is the high line of △ GCH, ∵ CHG is the isosceles triangle with Hg as the bottom, we can get DH = dg