國三圓的問題求解!謝謝! 1.梯形ABCD為圓O的外切四邊形，AB(左邊的邊常)=12，ㄥA=ㄥB=90度，梯形梯形ABCD的面積為150，則BC(下底)的長度=? 2.圓內接四邊形ABCD中，已知AB=16，BD=20，AD=12，ㄥADC=120度，則AC=?

1)

梯形ABCD面積 = 150
(AD + BC)AB/2 = 150
(AD + BC)12/2 = 150
AD + BC = 25

DC切點距D = AD切點距D = AD - 圓半徑 = AD - AB/2 = AD - 6 ,
DC切點距C = BC切點距C = BC - 圓半徑 = BC - 6
故 DC = AD - 6 + BC - 6 = AD + BC - 12 = 25 - 12 = 13

最後 AB² + (BC - AD)² = DC²
12² + (BC - (25 - BC))² = 13²
(2BC - 25)² - 25 = 0
(2BC - 30)(2BC - 20) = 0
BC = 15 或 BC = 10(捨因 BC > 直徑 = 12)

2)

因 16² + 12² = 20² ⇒ AB² + AD² = BD² , 故∠BAD 為直角, BD 為直徑。
而 ∠ADC = 120°, 則∠ABC = 180 - 120 = 60°, 故 AC 是圓內接正三角形之邊, 得 (AC/2) / 半徑 = cos30°,
(AC/2) / 10 = √3/2
AC = 10√3

不用三角函數解:
設圓心為O, 則反角AOC = 2∠ADC = 240°, 得∠AOC = 360 - 240 = 120°.
設AC中點為M , 則∠AMO為直角, ∠AOM = ∠AOC / 2 = 60° , 故 2OM = OA = 10,
OA² = OM² + AM²
10² = 5² + AM²
AM = 5√3
AC = 2AM = 10√3