Giving two equal line segments AC and BD so that they bisect each other at E. Prove that quadrilateral ABCD is a Rectangle.

Question

Given $AC=BD$

so $\angle AED= \angle CEB$ by Prop I.15

$AE+EC$ and $BE=DE$ by definition of bisect

So by $SAS$, $\triangle AED$ is congruent to $\triangle BEC$ (postulate 12) Therefor $AD=BC$

Then by Prop I. 15 $\angle AEB= \angle CED$ and $\triangle BEA \cong \triangle CED$

so $CD=AB$.

$\angle EDA= \angle EBA$, by Prop I.34

so $BC\parallel AD$

$\angle EAB= \angle ECD$ by Prop. I.34

so $BA=CD$

But how do I prove that the angles are right angles?

Answer 1

Hints:

1) A quadrangle is a parallelogram iff its diagonals bisect each other

2) A paralellogram is a rectangle iff its diagonal's lengths are equal

Answer 2

Some hints:

Let $M$ be he midpoint of $AD$. Since the triangles $\triangle(AEM)$ and $\triangle(DEM)$ have equal sides they are congruent. It follows that $\angle(AME)=\angle(DME)={\pi\over2}$, and that $E\vee M$ is an angle bisector at $E$.