解答

✨ 最佳解答 ✨

11.
等腰三角形,過A點作中垂線交BC於E點,BE = 9 => AE = 12
(1)
面積 = 1/2 × 18 × 12 = 108
(2)
面積 = 108 = 1/2 × AB × CD = 1/2 × 15 × CD => CD = 72/5

12.
假設 ∠ABF = ∠CBF = x,則
∠BAD = 90-2x => ∠CAD = 90-(90-2x) = 2x
∠AFB = 90-x
∠ACB = 90-2x
△AEF中,∠AEF = 180 - ∠CAD - ∠AFB
= 180 - 2x - (90-x) = 90-x
即 ∠AFB = ∠AEF = 90-x,故為等腰三角形。

13.
假設 ∠BAC = ∠BCA = x,則
∠BDC = ∠DBC = 2x => ∠BCD = 180-4x
∠DCE = ∠DEC = 180 - (180-4x) - x = 3x => ∠CDE = 180-6x
∠DFE = ∠FDE = 180 - (180-6x) - 2x = 4x => ∠DEF = 180-8x
又因為 ∠BAC = ∠DEF = x
180 - 8x = x, 9x = 180, x = 20度

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