∴BE+EC=CF+EC

∴BC=EF

In triangle ABC and triangle DEF,

AB=DE

BC=EF

AC=DF

∴ triangle ABC≌ triangle DEF(SSS)

∴∠B=∠DEF,∠ACB=∠DEF

∴AB‖DE,AC‖DF

9.∵∠BCE+∠ACD=90？ ，∠CAD+∠ACD=90？

∴∠BCE=∠CAD

You are ∵BE⊥CE,AD⊥CE.

∴∠E=∠ADC=90？

In triangle BCE and triangle CAD,

∠E=∠ADC

∠BCE=∠CAD

BC=CA

∴ triangle BCE≌ triangle CAD(AAS)

∴ce=ad=2.5cm,be=cd=ce-de=2.5- 1.7= 1.8cm

1 1. Let DE⊥AB be in E,DF⊥AC's F.

Proof: As shown in the figure, DE⊥AB is in E,DF⊥AC is in F.

AD is the angular bisector.

∴DE=DF

∴S triangle ABD? AB×DE AB

__________=________=__

S triangle ACD? AC×DF AC

Namely: s triangle ACD=AB:AC