五点共圆问题和密克点

由图可得，
点 D 是完全四边形 FE'ID'HC' 的密克点，
且点 B 是完全四边形 IA'FB'GC' 的密克点，
从而 FBC'I 四点共圆，
且 FC'DI 四点共圆，
所以 FBC'DI 五点共圆。

因为 FBC'DI 五点共圆，
所以 \(\angle\)FBC'+\(\angle\)E'ID' = 180\(^\circ\)

导角可得，
\(\angle\)FBC' + \(\angle\)E'ID = \(\angle\)FBA + \(\angle\)ABD + \(\angle\)DBC' + \(\angle\)E'ID'
=\(\angle\)FBA + \(\angle\)ABD + \(\angle\)E'ID
\(\angle\)FBA = \(\angle\)FA'A = \(\angle\)AJE'
\(\angle\)FBA + \(\angle\)ABD + \(\angle\)E'ID = \(\angle\)AJE' + \(\angle\)ABD + \(\angle\)E'ID
\(\angle\)FBC' + \(\angle\)E'ID = \(\angle\)AEE' + \(\angle\)ABD + \(\angle\)E'ED = \(\angle\)AED + \(\angle\)ABD = 180\(^\circ\)

对角互补，从而 ABDE 四点共圆。
类似可证任意四点共圆，从而 ABCDE 五点共圆。

\(\square\)