∵MD⊥AD,ME⊥DH,∴∠M=∠ADH=2∠ADE
∵tan∠ADE=1/2,∴tan∠M=4/3
(1/sinM)²=1+(tan∠M)²=25/16
∴1/sinM=5/4=MD/DH=(MC+CD)/DH=MC/DH+1
故MC/DH=1/4
∵MD⊥AD,ME⊥DH,∴∠M=∠ADH=2∠ADE
∵tan∠ADE=1/2,∴tan∠M=4/3
(1/sinM)²=1+(tan∠M)²=25/16
∴1/sinM=5/4=MD/DH=(MC+CD)/DH=MC/DH+1
故MC/DH=1/4