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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
1 year, 10 months 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.

- 1 year, 10 months ago

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