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標題:
發問:
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
最佳解答:
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)
其他解答:
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)
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Circle發問:
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
最佳解答:
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)
其他解答:
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)
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