, the area of ABC is 63, D is a point on BC, BD ∶ CD = 2 ∶ 1, DE∨AC intersects AB at point E, and the area of CDF extends from DE to F, so Fe ∶ ED = 2 ∶ 1.

Solution: MW⊥BC, AN⊥BC, and vertical feet are W and N respectively.

∫BD:CD = 2: 1，DE∑AC，

∴BE:AE=2: 1，

∴BD:BC=DE:AC=BE:AB=2:3，

∴S△BDE:S△ABC=4:9，

∴S△BDE=49×63=28，

∫FE:ED = 2: 1 = 4:2，

∴EF:AC=4:3，

∴S△MEF:S△AMC= 16:9，

∴EM:AM=4:3，

Suppose EM=4x, AM=3x, BE = 23ab = 2ae = 2 (EM+AM) =14x,

∴BM:AB= 18x:3x= 18:3，

∴mw:an=bm:ab= 18:2 1=6:7，

∴S△BMC:S△ABC= 12BC？ WM: 12BC？ AN=WM:AN=6:7，

∫S△ABC = 63，

∴S△BMC=54，

∴S△AMC=63-54=9，

∫S△MEF:S△AMC = 16:9，

∴S△MEF= 16，

∫S△BDE = 49×63 = 28，

∴S quadrilateral MEDC=63-9-28=26,

The area of CDF is: 26+ 16 = 42.

So the answer is: 42.