Giving two equal line segments AC and BD so that they bisect each other at E. Prove that quadrilateral ABCD is a Rectangle.
By Sebastian Wright
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?
$\endgroup$ 12 Answers
$\begingroup$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
$\endgroup$ $\begingroup$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 $E$.
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