Prove that ME=MC in a Circle Geometry ABCD

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SUMMARY

In the discussion, it is established that in cyclic quadrilateral ABCD, where diagonals AC and BD intersect at right angles at point E, the segment ME, where M is the midpoint of CD, is equal to segment MC. The proof relies on properties of cyclic quadrilaterals and the geometric relationships formed by the intersection of the diagonals. The conclusion confirms that ME = MC, reinforcing the symmetry inherent in cyclic quadrilaterals.

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ABCD is a cyclic quadrilateral. The diagonals AC and BD intersect at right angles at E. M is the midpoint of CD. ME produced meets AB at N.

Show that ME = MC
 

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Lukybear said:
ABCD is a cyclic quadrilateral. The diagonals AC and BD intersect at right angles at E. M is the midpoint of CD. ME produced meets AB at N.

Show that ME = MC

What are your thoughts on how to proceed?
 
Nvm, I've found solution. Thxs.
 

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