# Proving conditions for angles in two triangles with one equal angle

If in triangles $ABC$ and $DEF$, we have $\angle BAC = \angle EDF$ and $AB:DE = BC:EF$. Then prove that either $\angle ACB + \angle DFE = 180^\circ$ or $\angle ACB = \angle DFE$.

My attempt:

If $\angle ABC = \angle DEF$ then the two triangles are similar and hence $\angle ACB = \angle DFE$. If $\angle ABC \neq \angle DEF$ then let $\angle ABC > \angle DEF$. We have, $$\angle ACB + \angle DFE = \angle ACB + (\angle ACB + \angle ABC - \angle DEF)$$ $$= 180^\circ - \angle BAC + \angle ACB - \angle DEF$$ So it all boils down to proving that $\angle ACB = \angle BAC + \angle DEF$. How do I do that?

This is a diagram of the second case

• That's done to encase one triangle into another as essentially $\angle A = \angle D$. It can be used to get the answer. Mar 4, 2018 at 11:12

By the sin rule $$\frac{EF}{BC}=\frac{\frac{EF}{\sin \angle D}}{\frac{BC}{\sin \angle A}}= \frac{\frac{DE}{\sin \angle DFE}}{\frac{AB}{\sin \angle ACB}}.$$ Hence $$\sin \angle DFE=\sin \angle ACB$$ which is equivalent to the claim.