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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.

Note by Choi Chakfung
6 months, 4 weeks ago

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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 · 6 months, 2 weeks ago

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