[求助] 求助一道平面几何证明题已有2人参与

2楼: Originally posted by alober at 2017-06-26 08:54:25
\angle ABP=x,BC=a \Rightarrow BQ:BC=\sin30^\circ:\sin(60^\circ+x),AB:BC=\cos(30^\circ+x) \Rightarrow \frac{\tan\angle AQP}{\tan\angle x} = \frac{\tan\angle AQP}{\tan\angle ABP} = \frac{AP:AQ}{AP:AB}  ...

6楼: Originally posted by 一直朝前走 at 2017-06-26 22:07:19
中间如图应是平行线吧

16楼: Originally posted by k0waii at 2017-06-27 05:07:31
qc和pb的交点作底边平行线由此得出等边直角结论45度是15的三倍 结束

18楼: Originally posted by 李连玥 at 2017-06-27 07:54:31
∵ΔABC是等腰直角三角形，∴∠ABC=45°,∠ACB=45°。∵QP∥BC,∴∠ABC=∠AQP=45°（两直线平行同位角相等）,又∵∠PBC=30°,∴∠1=15°

21楼: Originally posted by tianliangzh at 2017-06-27 09:08:53
题上都给的什么条件？

42楼: Originally posted by hzh20070428 at 2017-07-25 11:23:58
已证明出来了，待整理中。涉及初中知识点：三角形相似、垂心、四点共圆。楼主还需要吗？

51楼: Originally posted by Mr__Right at 2017-08-05 08:59:22
这个漂亮不？？

http://blog.csdn.net/stereohomology/article/details/76697160...