∫△ABC, where BA=BC.

∴ ∠A=∠C

* df⊥ac

∴ ∠DFA=∠EFC=90

∴, ∠ D = 90-∠ A in Rt△DFA.

Similarly, in Rt△EFC, ∠ cef = 90-∠ c.

∴ ∠D=∠CEF

I went to bed again.

∴∠BED=∠D

△ DBE is an isosceles triangle.

2.( 1) Proof:

∵ △ABC and△△△ CDE are equilateral triangles.

∴ BC=AC，CE=CD，∠BCF=∠DCH=60

∴ ∠BCF+∠FCH=∠DCH+∠FCH

That is ∠BCE=∠ACD

∴△BCE?△ACD(SAS)

(2) prove that:

∫△BCE?△ACD

∴ ∠CBF=∠CAH

∠∠BCF+∠DCH+∠FCH = 180。

∠∠BCF =∠DCH = 60。

∴∠fch=60 =∠ ace

∴∠BCF=∠ACE

BC = AC

∴△BCF?△ace

∴ CF=CH

(3) prove that:

CF = CH

∴ △CFH is an isosceles triangle

∴ ∠CFH=∠CHF

And fch = 60, and < cfh+< CHF+< fch =180.

∴ ∠CFH=∠CHF=∠FCH=60

∴ △CFH is an equilateral triangle

3. Proof: Connect CE

∫△ABC is an equilateral triangle

∴ BC=AC，∠ACB=60

EA = EB，EC=EC。

∴△BCE?△ace(SSS)

∴ ∠BCE=∠ACE=( 1/2)∠ACB=30

In △CBE and △DBE.

Divide DBC equally

∴∠CBE=∠DBE

BD = AC

∴ BD=BC

Become again

∴△CBE?△DBE(SAS)

∴ brominated diphenyl ether = brominated diphenyl ether

∫∠BCE = 30。

∴ ∠BDE=30