# Given that $BA \parallel DE$, $AB = BC$, $CD = DE$, and $B$,$C$, $D$ are collinear, prove that $\angle ACE = 90^\circ$

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.

Triangles $CDE$ and $ABC$ are isosceles.

Angles $D$ and $B$ are supplementary

$2\hat{E}+\hat{D}+2\hat{C}+\hat{B}=360°$

$2\hat{E}+2\hat{C}=180°$

$\hat{E}+\hat{C}=90°$

Thus $\widehat{ACE}=180°-(\hat{E}+\hat{C})=180°-90°=90°$

$\widehat{ACE}=90°$

QED

Hope this helps

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we denote with $\angle A=\angle C$ and $\angle C=\angle E=\beta$ and $\angle B=x,\angle D=y$ then we get $$\alpha+\beta=180^{\circ}-\frac{x+y}{2}$$ and since $x+y=180^{\circ}$ we get $\alpha+\beta=90^{\circ}$