Prove that ME=MC in a Circle Geometry ABCD

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ABCD is established as a cyclic quadrilateral with diagonals AC and BD intersecting at right angles at point E. The midpoint M of side CD is identified, and line ME is extended to meet line AB at point N. The goal is to demonstrate that the length of ME equals the length of MC. A solution has been found, indicating that the problem can be resolved effectively. The discussion highlights the geometric properties of cyclic quadrilaterals and the relationships between their segments.
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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.
 

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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