∴∠A+∠B=90，

∫∠ACD =∠B，

∴∠A+∠ACD=90，

∴∠ADC=90，

Namely CD⊥AB,

In the proof, "two acute angles of a right triangle are complementary" and "two triangles with complementary acute angles are right triangles" are applied.

(2) Proof: ∫AE bisection ∠BAC,

∴∠CAE=∠BAE，

∠∠CAE+∠AEC = 90，∠BAE+∠AFD=90，

∴∠AEC=∠AFD，

∠∠AFD =∠CFE (equal to the vertex angle),

∴∠aec=∠cfe；

(3) Solution: BC = 3CE, AB=4AD,

∴s△acd= 1/4s△abc= 1/4×36=9,s△ace= 1/3s△abc= 1/3×36= 12，

∴S△CEF-S△ADF=S△ACE-S△ACD

= 12-9

=3.

So the answer is: 3.

The fastest answer, hope to adopt, thank you.