https://i.na.cx/93b0m1.png
1. In the figure, chord BA and chord DC are produced to meet at P. If PA = 4, AB = 5 and PC = 5, then CD = ?
https://i.na.cx/Voe6qa.png
2. DB tangent to the circle, AC = 4, CD = 2. Find AB which is diameter
Circle
2014-09-17 3:03 pm
回答 (2)
2014-09-17 5:10 pm
✔ 最佳答案
1. In Δs PAC, PDB∠PAC＝∠PDB ⋯⋯ (ext.＝int. opp., cyclic quad.)
∠PCA＝∠PBD ⋯⋯ (ext.＝int. opp., cyclic quad.)
∴ ΔPAC～ΔPDB ⋯ (AAA)
ie. PA/PD＝PC/PB
==> 4/(5＋CD)＝5/(4＋5)
==> CD＝4*9/5－5
∴ CD＝2.2 (units)
2. Join BC, as AB is a diameter, so
∠ACB＝90° ⋯⋯⋯⋯ (angle in semi-circle)
And, DB is a tangle, so
∠ABD＝90° ⋯⋯⋯⋯ (tangent ⊥ radius)
In Δs ACB, ABD
∠A＝∠A ⋯⋯⋯⋯⋯ (common)
∠ACB＝∠ABD ⋯⋯ (＝90°, proved)
∴ ΔACB～ΔABD ⋯ (AAA)
ie. AC/AB＝AB/AD
==> AB²＝AC*AD＝4*(4＋2)＝24
∴ AB＝2√6 (units)
2014-09-17 9:06 pm
1. Alternative method :
PA × PB = PC × PD (intersecting chord)
4 × (4 + 5) = 5 × (5 + CD)
5 + CD = 7.2
CD = 2.2 ...... (ans)
2014-09-17 13:13:50 補充：
2. Alternative method :
AB² = AD² - BD² (Pythagorean theorem)
But BD² = CD × DA (intersecting chords)
Hence, AB² = AD² - CD × DA
AB² = (2 + 4)² - 2 × (2 + 4)
AB = 2√6 ...... (ans)
PA × PB = PC × PD (intersecting chord)
4 × (4 + 5) = 5 × (5 + CD)
5 + CD = 7.2
CD = 2.2 ...... (ans)
2014-09-17 13:13:50 補充：
2. Alternative method :
AB² = AD² - BD² (Pythagorean theorem)
But BD² = CD × DA (intersecting chords)
Hence, AB² = AD² - CD × DA
AB² = (2 + 4)² - 2 × (2 + 4)
AB = 2√6 ...... (ans)
收錄日期: 2021-05-02 11:04:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140917000051KK00011