2013年河南中考数学试题第23题第(3)问,命题思路独辟蹊径,把特殊锐角45°引入,探究以直线与抛物线在y轴上的交点为顶点,经过抛物线上一动点P的直线与已知直线夹角为45°,由于角度特殊,构造三角形即是等腰直角三角形,两条直角边长度相等,联系直角三角形性质、三角形相似、勾股定理等可以确定有关线段长度,下面结合问题,探究第三问的几种解法,供参考.例(河南中考数学试题第23题)如图1,抛物线y=-x2+bx+c与直线y=12x+2交于C、D两点,其中点C在y轴上,点D In 2013 Henan senior high school entrance examination mathematics questions 23 (3) Q, propositional thinking of a new approach, the special acute angle of 45 ° introduced to explore the straight line and the parabola on the y-axis of the vertex, Because of the special angle, the constructed triangle is the isosceles right-angled triangle, the two right-angled sides are equal in length, and the relationship between the right triangle, the triangle, the Pythagorean theorem and the like can determine the length of the line segment, Example (Henan test math test question 23) Figure 1, the parabola y = -x2 + bx + c and the line y = 12x +2 to pay C, D two points, Where point C is on the y-axis, point D