90 is not equal 90

Given that $AB=DC,DA$ perpendicular to $AB$ . Let M and N be the perpendicular bisector of $AD$ and $BC$ respectively. These two perpendicular bisector intersect in $E$ . Joining $AE,ED,EB,EC$ since $EM$ is perpendicular bisector $AE=ED$ since $EN$ is perpendicular bisector $EB=EC$ $\triangle EAB=\triangle EDC$ (SSS) $\angle EAB=\angle EDC$ $\angle EAD=\angle EDA$ we have $\angle MAB=\angle MDC$ obviously it is impossible.

#Geometry

Easy Math Editor

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Comments

Here's an accurate graphic of how it should look like

We can see that $\Delta ABE$ is congruent with $\Delta DCE$ , and the paradox then vanishes.

Michael Mendrin - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$