I hope someone can point me in the right direction on the following question.
Given: $BA \parallel DE$, $AB = BC$, $CD = DE$, and $B$, $C$, $D$ are collinear.
Prove: $\angle ACE = 90^\circ$.
I've determined that $\angle ACE$ does not equal to $90^\circ$. In the diagram there's two triangles which are $\triangle ABC$ and $\triangle CDE$. Both of them are isosceles triangles, because $AB = BC$ and $CD = DE$.
The reason why I'm assuming that $\angle ACE$ does not equal to $90^\circ$ is because $\triangle ABC$ and $\triangle CDE$ is not right angled triangle, because in the question or the diagram it does not state that:
- Line $AB$ is perpendicular to line $BC$
- Line $CD$ is perpendicular to line $DE$
If they were perpendicular, then it would make sense that $\angle ACB$ and $\angle ECD$ would equal to $45^\circ$. Since $B$, $C$, $D$ is collinear, I can subtract angles $\angle ACB$ and $\angle ECD$ from $180^\circ$ and get $90^\circ$.
But I also think I might be wrong because $AB\parallel DE$, which would make:
- $AB$ perpendicular to $BC$
- $DE$ perpendicular to $CD$
Thank you for time.