∠ABC=90°点o是AC的中点,点P是斜边AC上的一动点

显示全部楼层 · 2012-10-8 11:27:56

1.证明△POB 和 △DEP 全等即可证明第一题∠POB = ∠DEP = 90°PB = DP∠PBD = ∠ PDB∠PBD = ∠PBO + ∠OBC =∠PBO + 45°∠PDB = ∠DPC + ∠C = ∠DPE + 45°所以∠DPE = ∠PBO ，所以 △POB ≌ △DEP2. AC =2 , 所以AB = BC = √2 ，BO = OA = OC = 1OP = DE = EC = 1 - x S△ABC = 1S△PBA = AP * OB / 2 = x/2S△DEC = DE*EC / 2 = (1-x)2/2y=S△ABC-S△PBA-S△DEC= 1...

千问 · 2012-10-8 11:27:56

(1)用解析几何做以B为原点BC为x轴正向建立直角坐标系设AB=AC=√2D(Xo,0)由于PB=PD则P点的横坐标为BD的中点则P的横坐标为Xo/2 AC的直线方程为X+Y=√2(a)将横坐标代入得P(Xo/2,√2-Xo/2)DE的直线方程为Y/(X-Xo)=√2
(b)联立方程（a)(b)得E的坐标...