In the accompanying figure, ΔABCis given with $\displaystyle \overline{AB}$ $\displaystyle \overline{AC}$. Can you construct a proof based on the diagram shown, that ∠ABC ∠ACB? (This was Euclid'spons asinorum).

Now I tried looking up pons asinorum proofs online but couldn't quite find one that made sense to me.

Here is my work so far:

Know: m∠BDE = m∠CDF (By the vertical pair theorem that vertical angles have equal measures)

$\displaystyle \overline{AB}$ $\displaystyle \overline{AC}$

$\displaystyle \overline{BE}$ $\displaystyle \overline{CF}$

$\displaystyle \overline{AE}$ = $\displaystyle \overline{AB}$ + $\displaystyle \overline{BE}$

$\displaystyle \overline{AF}$ = $\displaystyle \overline{AC}$ + $\displaystyle \overline{CF}$

$\displaystyle \overline{AE}$ = $\displaystyle \overline{AF}$

ΔABCis isosceles, so isn't m∠ABC = m∠ACB by the Isosceles Triangle Theorem? And thus ∠ABC ∠ACB? Or is that what I was supposed to prove in the first place?

I am quite confused here

Any help is greatly appreciated.

Thank you very much for your time.