∴∠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.