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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
By connecting CG, ∵ ad ⊥ BC, ∵ ABC + ⊥ gab = 90 °, we can get ∵ ABC + ⊥ FCB = 90 ° and ∵ gab = ∵ GCB is the same arc BG, ∵ gab = ∵ GCB, we can get ∵ GCB = ∵ FCB, ∵ CD ⊥ GH, that is, CD is the high line of △ GCH, and ∵ CHG is the bottom of Hg
As shown in the figure, in △ ABC, ad, be and CF are three high lines, the intersection point is h, extend the circumscribed circle of ah intersection at point m, (1) prove: ∠ FHB = ∠ BAC; (2) try to guess the quantitative relationship between line DH and line DM, and prove your conclusion
(1) It is proved that: ∵ be, CF is the height of △ ABC, ∵ AEB = ∵ CFB = 90 °, ∵ Abe + ∵ BAE = 90 °, ∵ FBH + ∵ FHB = 90 °, ∵ FHB = ∵ BAC; (2) DH = DM. The reasons are as follows: connecting BM, ∵ ad, be is the height of △ ABC, ∵ BEC = 90 °, ∵ BDA = 90 °, CBE =
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