Question

如图,在矩形ABCD中,对角线BD的垂直平分线MN与AD相交于点M,与BC相交于点N,连接 BM,DN．

Answer 1

BD=√﹙4²+8²﹚=4√5

OD=1/2BD=2√5

∵∠DOM=90º=∠A,∠ADB=∠ODM

∴△ABD∽△OMD

∴AD:OD=BD:MD

∴MD=﹙2√5·4√5﹚/8=5