F5 tangent of circle20mark
2011-01-16 5:51 am
圖片參考：http://imgcld.yimg.com/8/n/HA00763813/o/701101150128913873369820.jpg
In the figure, AC touches the two circles at A and C respectively. B is a point on AC and BD is the common tangent to the two circles at D.Prove that
a) AB = BC
b) ∠ADC =90°
回答 (2)
2011-01-16 6:01 am
✔ 最佳答案
a)AB = BD (tangent properties) BD = BC (tangent properties) So AB = BCb)ㄥBAD = ㄥBDA (base ∠s, isos. Δ)ㄥBDC = ㄥBCD (base ∠s, isos. Δ)On the other hand ,ㄥBAD + (ㄥBDA + ㄥBDC) + ㄥBCD = 180° (ㄥ sum of △)(ㄥBAD + ㄥBDA) + (ㄥBDC + ㄥBCD) = 180°2ㄥBDA + 2ㄥBDC = 180°ㄥBDA + ㄥBDC = 90°ㄥADC =90°
2011-01-16 6:03 am
(a) Consider the small circle
AB = BD (tangent properties)
Then consider the large circle
BC = BD (tangent properties)
So AB = BC
(b)
∠ BAD = ∠ BDA and ∠ BCD = ∠ BDC
Since ∠ BAD + ∠ BDA + ∠ BCD + ∠ BDC = 180
2 ( ∠ BDA + ∠ BDC ) = 180
∠ BDA + ∠ BDC = 90
∠ ADC = 90
AB = BD (tangent properties)
Then consider the large circle
BC = BD (tangent properties)
So AB = BC
(b)
∠ BAD = ∠ BDA and ∠ BCD = ∠ BDC
Since ∠ BAD + ∠ BDA + ∠ BCD + ∠ BDC = 180
2 ( ∠ BDA + ∠ BDC ) = 180
∠ BDA + ∠ BDC = 90
∠ ADC = 90
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