已知四边形ABCD中AB=AC=BD且AC垂直于BD,垂足为O.I为三角形ABO内心.M为AB中点.求证：MI垂直于CD且MI=1/2CD.

Question

已知四边形ABCD中AB=AC=BD且AC垂直于BD,垂足为O.I为三角形ABO内心.
M为AB中点.求证：MI垂直于CD且MI=1/2CD.

Answer 1

设△ABE的内切圆切AB于N,切BE于P,切EA于Q.∵AC⊥BD于E,∴设AN=AQ=x,BN=BP=y,EP=EQ=IN=r.由AE^2+BE^2=AB^2,得 (x+r)^2+(y+r)^2=(x+y)^2,∴r(x+y)=xy-r^2.而MN=|AM-AN|=|(x+y)/2-x|=|y-x|/2,IN⊥AB,∴IM^2=IN^2+MN^2=r...