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(1999年山东省初中数学竞赛)如图1,AD是Rt△ABC斜边BC上的高,P是AD的中点,连结BP并延长交AC于E,已知AC：AB=R．求AE：EC．分析：由已知AC：AB=R,可求出BD：DC的值．根据Rt△ABD∽Rt△CBA,Rt△CAD∽Rt△CBA,可得AB2=BD·BC,AC~2=DC·BC,从而求得(BD)/(DC)=(AB~2)/(AC~2)=1/R~2,所以(BD)/(BC)=1/(1+R~2),然后再求AE：CE的值．我们知道要求比值,一般需借助于平行线,
(1999 Shandong Junior High School Mathematical Competition) As shown in Figure 1, AD is the height of BC on the hypotenuse of Rt △ ABC, P is the midpoint of AD, links BP and extends AC to E, known AC:AB=R. Seeking AE:EC. Analysis: From the known AC:AB=R, the value of BD:DC can be obtained. According to Rt △ ABD ∽ Rt △ CBA, Rt △ CAD ∽ Rt △ CBA, available AB2 = BD · BC, AC ~ 2 = DC · BC, in order to obtain (BD) / (DC) = (AB ~ 2) / (AC~2)=1/R~2, so (BD)/(BC)=1/(1+R~2), and then find the value of AE:CE. We know that we need a ratio, generally we need parallel lines.