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發問:
In the figure, AB is a diameter of the circle.AC is a tangent to the circle at A. BDC is a straight line. Show that▲ABC~▲DBA. 圖: http://hk.geocities.com/yokikwok2003/tangent.bmp
最佳解答:
圖片參考：http://hk.geocities.com/yokikwok2003/tangent.bmp In ▲ABC and ▲DBA, angle BAC = 90° (tangent ┴ radius) angle BDA = 90° (angle in semicircle) Thus angle BAC = angle BDA...................(1) angle ABC = angle DBA (common angle)...............(2) angle BCA = 180° - angle CAB - angle ABC (angle sum of ▲) = 180° - angle ADB - angle DBA (proved) = angle BAD (angle sum of ▲)..........................(3) Therefore ▲ABC ~ ▲DBA (AAA)
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F.5 deductive geometry--tangets to circle發問:
In the figure, AB is a diameter of the circle.AC is a tangent to the circle at A. BDC is a straight line. Show that▲ABC~▲DBA. 圖: http://hk.geocities.com/yokikwok2003/tangent.bmp
最佳解答:
圖片參考：http://hk.geocities.com/yokikwok2003/tangent.bmp In ▲ABC and ▲DBA, angle BAC = 90° (tangent ┴ radius) angle BDA = 90° (angle in semicircle) Thus angle BAC = angle BDA...................(1) angle ABC = angle DBA (common angle)...............(2) angle BCA = 180° - angle CAB - angle ABC (angle sum of ▲) = 180° - angle ADB - angle DBA (proved) = angle BAD (angle sum of ▲)..........................(3) Therefore ▲ABC ~ ▲DBA (AAA)
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