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A、B、c是⊙0上的三点,以BC为一边,作∠CBD=∠ABC,过BC上一点P,作PE‖AB交BD于点E．若圆心角∠AOC=60°,BE=3,则点P到弦AB的距离为?
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A、B、c是⊙0上的三点,以BC为一边,作∠CBD=∠ABC,过BC上一点P,作PE‖AB交BD于点E．若圆心角∠AOC=60°,BE=3,则点P到弦AB的距离为?
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答案和解析
过点E做EF垂直AB于F 过p做PG垂直AB于G,因为∠CBD=∠ABC=1/2∠AOC=30度,所以∠ABD=60度 在直角三角形EBF中EF=3倍（根号3）/ 2 由于平行和垂直 可证PGFE为矩形 所以EF就是p到AB的距离
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