Δ KAC ∼ Δ KBD
KA: KB=AC:CD
KA: KB=40:60
KA:KB=2:3 ⇒ 3KA=2KB
KB=KA+AB
AB=KB-KA=KB-(2/3)KB=(1/3)KB
AB=AM+MB
AM:MB=6:14
AM=6k
MB=14 k
[b]AB=20k[/b]
AB=(1/3)KB
KB=60k
KA=(2/3) KB=40k
Δ KAC ∼ Δ KMN
KA:AC=KM:MN
40k: 40=46k:MN
[b]MN=46[/b]