为避免打字中大量中英文切换，我用英文写出。

We first state a theorem. In ∆ABC, if AD (D on BC) bisects ∠A, then AB/AC = AD/CD.

Our problem is this. In ∆ABC, AD and AE trisect ∠A. Here D and E are on BC with the order from left B, D, E, and C. Is it possible that BD = 1, DE = 2, and EC = 4?

∠BAD = ∠DAE = ∠EAC = a (Given).

AE/AB = 2, AC/AD = 2 (Apply theorem).

Take E’ on AE so that ∠ADE' = ∠ABD.

We get immediately

∆ABD ~ ∆ADE'

 Here “~” means similar (may not be same).

∠ADE = ∠ABD + a (exterior ∠ of ∆ABD).

Because ∠ADE' = ∠ABD,

∠E’DE = ∠ADE – ∠ADE’ = a.

Look at ∆ABD and ∆DE’E. ∠DAE = ∠E’DE =a. They both have ∠AED.

∆ADE ~ ∆DE'E.

AE/DE = DE/E’E è EE’ = (DE)^2/AE = 4/AE.

Using ∆ABD ~ ∆ADE', we have

EE’ = AE – AE’ = AE – (AD)^2/AB.

Eliminating EE’, using AE/AB = 2,

4 = (AE)^2 – 2(AD)^2.

Similarly, working on ∆ADE and ∆AEC,

16 = (AC)^2 – 2(AE)^2.

Eliminating AE, we have

24 = (AC)^2 – 4(AD)^2 = 0.

原来以为我哪儿出错了，看来确实无解。