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?
Best Answer
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