论文部分内容阅读
1991年9月号问题解答 (解答由供题人给出) 7.在Rt△ABC中,AD为斜边BC上的高,在AB、AC上各取一点M、N,满足DM⊥DN。试证:△BDM与△CDN的外接圆外切且直线MN是这两圆的一条公切线。证明易知A、M、D、N四点共圆,可得∠DMN=∠DAN=∠ABC。 (1)若∠BMD=90°,则∠DNC=90°(如图1)。Rt△BDM、Rt△CDN的外心各是BD、DC的中点O_1、O_2,连结O_1M、O_2N,易证MN⊥O_1,M、MN⊥O_2N。此时既易证明△BDM与△CDN的外接圆外切,又不难证得直线MN是这两圆的
September 1991 question number (answer is given by the subject) 7. In Rt △ ABC, AD is the height on the hypotenuse BC, and each point AB and AC is taken as M, N, which satisfies DM ⊥ DN. Testimony: The circumscribed circles of ΔBDM and △CDN are circumscribed and the straight line MN is a common tangent of the two circles. Prove that the A, M, D, N four-point co-circle, available ∠ DMN = ∠ DAN = ∠ ABC. (1) If ∠ BMD = 90°, then ∠ DNC = 90° (Figure 1). The centers of Rt△BDM and Rt△CDN are O_1 and O_2 at the midpoints of BD and DC respectively, linking O_1M and O_2N, and MN⊥O_1, M, and MN⊥O_2N. At this point, it is easy to prove that the circumcircle of △BDM and △CDN is circumscribed, and it is not difficult to prove that the straight line MN is the two circles.