Please help me understand the logic/rule/method here. I can’t figure out how we can show that angle ACD and EDB are equal and angles CAB and DEB are equal in the attached problem. My issue is that I don’t see how it can’t be equally possible that ACD=DEB and CAB=EDB instead.
It follows from definition
\triangle ABC \sim \triangle EBD implies \angle A = \angle E, \angle B = \angle B, and \angle C = \angle D
Sorry I’m still trying to understand.
So if I were to take this same graphic and shifted point D up one unit, I could assume ∠A=∠E, and ∠C=∠D, because ∠B=∠B based on the rule that “△ABC∼△EBD implies ∠A=∠E,∠B=∠B, and ∠C=∠D”?
The vertex ordering is the tell-tale to identifying the corresponding and equal angles in two similar or congruent triangles. The statement \triangle ABC \sim \triangle EBD implies the following:
It’s the exact same logic for your question above.
I don’t think this is relevant to our discussion on how to parse “\triangle ABC and \triangle EBD are similar”
ah! Thank you!