Proving Miquel's Theorem: Need Help!

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SUMMARY

Miquel's Theorem states that for any triangle ABC, if points D, E, and F are selected within the sides BC, AC, and AB respectively, the circumcircles of triangles AEF, BDF, and CDE intersect at a single point M. To prove this, one effective approach involves identifying point M as the intersection of two circumcircles, then drawing lines MD, ME, and MF. Utilizing properties of cyclic quadrilaterals confirms that point M also lies on the third circumcircle.

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  • Basic knowledge of triangle geometry
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  • Study the properties of circumcircles in triangle geometry
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Mathematicians, geometry students, and educators seeking to understand or teach Miquel's Theorem and its applications in triangle geometry.

pholee95
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I don't know how to start proving this theorem, so can someone please help? I need to prove that the circumcircles all intersect at a point M. Thank you!

Miquel's Theorem: If triangleABC is any triangle, and points D, E, F are chosen in the interiors of the sides BC, AC, and AB, respectively, then the circumcircles for triangleAEF, triangleBDF, and triangleCDE intersect in a point M.

I have attached here the figure of theorem.
 

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pholee95 said:
I don't know how to start proving this theorem, so can someone please help? I need to prove that the circumcircles all intersect at a point M. Thank you!

Miquel's Theorem: If triangleABC is any triangle, and points D, E, F are chosen in the interiors of the sides BC, AC, and AB, respectively, then the circumcircles for triangleAEF, triangleBDF, and triangleCDE intersect in a point M.

I have attached here the figure of theorem.
Have you looked at https://en.wikipedia.org/wiki/Miquel's_theorem? The trick seems to be to take $M$ to be the point where two of the three circles meet, draw the lines $MD$, $ME$ and $MF$, then use properties of cyclic quadrilaterals to show that $M$ also lies on the third circle.
 

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