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

  • $\begingroup$ That's done to encase one triangle into another as essentially $\angle A = \angle D$. It can be used to get the answer. $\endgroup$ – Helix Mar 4 '18 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.


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