How do I solve this?

Question

tkhunny · Answer

Step Number 1 would be writing an actual question.
Stem Number 2 would be showing your best work.

anonymous · Answer

Sorry I thought it posted

tkhunny · Answer

Based on the tabs at the top of your screen, you might want to pay better attention to your studies!  :-)

You didn't do Step #2!!

tkhunny · Answer

The two triangles share angle D.  They each have a 60º angle.  If we can get the 3rd angle, we're done.

anonymous · Answer

I know what a two column proof is but I don't know how to do it with a triangle

anonymous · Answer

So how do we get the third one?

tkhunny · Answer

You have to decide which are the corresponding angles.

We're talking $\triangle FDE$ vs. $\triangle EDG$.

We have $\angle F \cong \angle E$ because they both measure 60º.
We have $\angle D \cong \angle D$ because they are the same angle.
We need $\angle E \cong \angle G$.

anonymous · Answer

I think its FEG and GDE

tkhunny · Answer

Sorry, I ordered them differently than the problem statement, but that doesn't change the task at hand.

What is the sum of the measures of the three angles of any triangle?

anonymous · Answer

180?

tkhunny · Answer

Definitely not FEG or GDE.  The problem statement says DEF and DGE.

anonymous · Answer

I honestly just don't understand any of what I'm doing or what you're saying

tkhunny · Answer

Okay, what is $m\angle E\;from\;\triangle FDE$?

Similarly, what is $m\angle G\;from\;\triangle EDG$?

Subtracting the two known congruent angles from 180º in each case provide the proof we need that the third angle is congruent.

anonymous · Answer

so i add the two 60s to make 120?

tkhunny · Answer

Sometimes, it helps quite a bit to draw the triangles NOT overlapping.  Can you redraw separately, $\triangle$FDE and $\triangle$EDG?

anonymous · Answer

and I subtract 120 from 180?

tkhunny · Answer

No.  No adding angles from different triangles.

$\triangle FDE$  $m\angle F + m\angle D + m\angle E = 180º$
$\triangle EDG$  $m\angle E + m\angle D + m\angle G = 180º$

$\triangle FDE$  $m\angle E = 180º - m\angle F - m\angle D$
$\triangle EDG$  $m\angle G = 180º - m\angle E - m\angle D$

Since we know $m\angle F = m\angle E$, we are done and $\angle E \cong \angle G$ and $\triangle FDE \sim \triangle EDG\;by\;AAA$.