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圆中有一类线段的倒数关系的证明,由于运用的知识点综合性强,往往难以形成解题的思路。本文通过搜集资料把这类问题的证明归纳为如下方法。1 利用勾股定理 例1AB是半圆的直径,⊙Q与半圆O内切于点P,与AB切于点D,设AD=a,DB=b,⊙Q的半径为r。求证:1/a+1/b=1/r。 证明如图1,连接QD,则△ODQ为直角三角形。 所以OQ2=OD2+DQ2,而OQ=1/2(a+b)-r,OD=1/2(a-b),
The proof of the reciprocal relationship of one type of line segment in a circle is often difficult to form a problem-solving idea because of the comprehensiveness of knowledge points used. This article summarizes the proof of this type of problem by collecting data as follows. 1 Using the Pythagorean Theorem Example 1AB is the diameter of a semicircle, ⊙Q and semicircle O are inscribed at point P, and AB is cut at point D. Let AD = a, DB = b, and the radius of ⊙Q be r. Verification: 1/a+1/b=1/r. Prove that in Figure 1, connecting QD, △ ODQ is a right triangle. So OQ2=OD2+DQ2, and OQ=1/2(a+b)-r, OD=1/2(a-b),