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In rectangle ABCD, diagonals AC and BD meet at E. Angles BAC and DCA are alternate (since AB and DC are parallel) and are therefore equal. The same is true of angles ABD and CDB. Also, AB = DC, so that triangles ABE and CDE are congruent. Thus, |AE| = |EC| and |BE| = |ED|, that is, the point E bisects both AC and BD.

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Q: How to Prove the diagonals of a rectangle bisect each other?
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