# Thread: Geometry Question: Proving two angles congruent

1. ## Geometry Question: Proving two angles congruent

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

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

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

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

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

ΔABC is 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.

2. Use the law of sines...

3. just drop a perpendicular from A on base BC.

try to prove it using RHS CONGRUENCY

4. ABC is a triangle with Ab = AC.
So you can simply use isosceles triangle approac (drop perpendicular on BC from A).
there is no need of points D,E and F.

But still if you want to use points D, E and F, here is approach.

first show that triangle EBC and triangle FCB are congurant
then angle EBC = angle FCB.
=> 180 - andge EBC = 180 - angle FCB.
=> angle ABC = angle ACB