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笔者所在学校初三期中试卷中有如下一题:在等腰直角△ABC中,∠BAC=90°,AB=AC,直线MN过点A且MN∥BC,以点B为一锐角顶点作Rt△BDE,∠BDE=90°,且点D在直线MN上(不与点A重合),如图1,DE与AC交于点P,易证:BD=DP.(1)在图2中,DE与CA延长线交于点P,BD=DP是否成立?如果成立,请给予证明;如果不成立,请说明理由.(2)在图3中,DE与AC延长线交于点P,BD与DP是否相等?请直接写出你的结论,无需证明.
I was in the school the first three papers in the paper has the following questions: △ ABC in the isosceles right angle, ∠ BAC = 90 °, AB = AC, the line MN over point A and MN ∥ BC, the point B as an acute angle vertices for Rt △ BDE, ∠BDE = 90 °, and the point D in the line MN (not coincident with the point A), as shown in Figure 1, DE and AC to point P, easy to prove: BD = DP. , DE and CA to extend the line to point P, BD = DP is established? If yes, please give proof; if not, please explain the reason. (2) In Figure 3, DE and AC extended line to point P, BD Is it equal to DP? Please write down your conclusion without proof.