初三数学题,急用,在线等在正方形ABCD中,P是CD边上一动点(与C、D不重合）,使三角尺的直角顶点与P点重合,并且一条

当∠EBP=∠PBC时,直角△PBE∽直角△CPB,(AAA);
且当∠EBP=∠PBC=α时,有∠DPE=∠PBC=α,直角△DEP∽直角△CPB,(AAA);
设BP=X,正方形ABCD边长=a,
tanα=(a-X*tanαcosα)/a...(1)
a²+(a-X*tanαcosα)²=X²...(2)
解方程组,由(1):
X=a(1-tanα)/(tanαcosα),
代入(2):
(tanα)²(cosα)²+(tanα)^4(cosα)²=1-2tanα+(tanα)²,[(tanα)^4表示4次方]
tanα=sinα/(cosα),代入上式,
(sinα)²+(sinα)^4/(cosα)²=1-2sinα/(cosα)+(sinα)²/(cosα)²,
(sinα)²(cosα)²+(sinα)^4=(cosα)²-2sinαcosα+(sinα)²,
(sinα)²[(cosα)²+(sinα)²]=(cosα)²-2sinαcosα+(sinα)²,
(sinα)²[1]=(cosα)²-2sinαcosα+(sinα)²,
2sinαcosα=(cosα)²
sinα/cosα=1/2,
tanα=1/2,
PC=BC*tanα=a/2,
DP=DC-PC=a-a/2=a/2,
AE=AD-DE=a-DP*tanα=a-(a/2)*1/2=3a/4,
tan∠ABE=AE/AB=(3a/4)/a=3/4,
[三角函数不熟练,解方程过程可能有更简便方法]

CD的中点

看不清题！。。。。......