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As shown in the figure, in △ ABC, ∠ C = 90 °, AC = BC, Da bisects ∠ cab and intersects BC at point D. can we determine a point E on AB so that the perimeter of △ BDE equals the length of AB? If yes, please point E and give proof; if not, please explain the reason
We can determine a point E on AB such that the perimeter of △ BDE is equal to the length of ab. it is proved that in △ ABC, ∠ C = 90 ° Da bisects ∠ cab, ｜ DC = De, ∠ CDA = ∠ EDA, ｜ AE = AC, ∵ AC = BC, ｜ ∠ B = 45 ° BC = AE, ｜ △ bed is isosceles right triangle, ｜ de = be
As shown in the figure, ∠ BAC = 90 ° in △ ABC, ad ⊥ BC at point D, CE bisects ∠ ACB, intersects ad at g, GF ∥ BC intersects AB at F, the verification is AE = BF
As shown in the figure, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\inthis paper, we present a new method for the determination of ACG CG (AAS), ∵ Ag = GH, ∵ BF = AG. ∵ age = ∵ CGD, in RT △ CDG, ∵ CGD + ∵ GCD = 90 °, ∵ age + ∵ GCD = 90 °. In RT △ CAE, ∵ ACG + ∵ AEG = 90 °, ∵ age = ∵ AEG, ∵ AE = AG, ∵ be = AG, ∵ AE = BF
Ad is the middle line of triangle ABC, e is the middle point of AD, the extension line of be intersects AC and F, and dg / / be intersects AC and G through point D
In Δ ADG:
∵ BF ∥ DG, e is the midpoint of AD,
∴AF/FG=AE/DE=1,
∴AF=FG,
In Δ BFC,
∵ D is the midpoint of BC,
∴CF、FG=CD/BD=1,
∴CF=FG,
∴AF=FG=CF.
As shown in the figure, ad bisects ∠ BAC, ad is parallel to eg, try to prove that △ AGF is isosceles triangle
Five minutes. Thank you
Because ad is parallel to eg, so ∠ GFA = ∠ bad (internal stagger angle) ∠ FGA = ∠ DAC (same position angle). And ad bisects ∠ BAC. So ∠ bad = ∠ DAC. So ∠ GFA = ∠ FGA. So triangle AGF is isosceles triangle
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