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20.[朝霞原创] (9分)如图,已知$\triangle ABC$是⊙O的内接三角形,$\overset{\frown}{AB} = \overset{\frown}{BC} = \overset{\frown}{AC}$,E是BC上的一点,连接AE,过点B作$BD// AE$交⊙O于点D,连接CD交AB于点F.
(1)求证:$AF = BE$;
(2)若$\angle CAE = 15^{\circ}$,请仅用无刻度的直尺在图中作出一个⊙O的内接等腰直角三角形(保留作图痕迹,不写作法).
(1)求证:$AF = BE$;
(2)若$\angle CAE = 15^{\circ}$,请仅用无刻度的直尺在图中作出一个⊙O的内接等腰直角三角形(保留作图痕迹,不写作法).
答案
20.解:(1)证明:
∵△ABC是⊙O的内接三角形,AB = BC = AC,
∴AB = BC = AC,∠ABC = ∠ACB = ∠BAC = 60°.(3分)
∵BD//AE,
∴∠ABD = ∠BAE.
∵∠ABD = ∠ACF,
∴∠ACF = ∠BAE.
∴△ACF≌△BAE.
∴AF = BE.(6分)
(2)在⊙O中作出满足条件的△BGH如图所示.
(答案不唯一)
