论文部分内容阅读
在解有关四边形的问题时.常常需要对问题分情况进行讨论.现归纳几例.供同学们学习时参考.例1(2011年·安顺)已知O为坐标原点,四边形OA BC为矩形,A(10,0),C(0,4).点D是OA的中点.点p在BC上运动.当△POD是腰长为5的等腰三角形时.点P的坐标为.分析:由于点P是一个动点,所以为使△POD是腰长为5的等腰三角形.需分情况讨论.解:根据已知得OA=10.D为OA的中点,则OD=5.使△POD为等腰三角形的情况有以下三种:(1)如图1,当P_1O=5时,因OC=4,由勾股定理,得P_1C=3,因此,点P_1的坐标为(3,4);
In the solution of the problem of the quadrilateral, often need to discuss the situation of the problem points. Now summarized a few cases for students to learn reference .Example 1 (Anshun in 2011) is known as the origin of coordinates O, quadrilateral OABC rectangular, A (10,0), C (0,4). Point D is the midpoint of OA. Point p moves on BC. When △ POD is an isosceles triangle with a waist length of 5. The coordinates of point P are. : Since point P is a moving point, so △ POD waist length is 5 isosceles triangle. Need to be divided into discussions. Solution: According to the known OA = 10. D is the midpoint of OA, OD = 5 (1) As shown in Figure 1, when P_1O = 5, due to OC = 4, the Pythagorean theorem, so P_1C = 3, so the coordinates of the point P_1 is (3,4);