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Simson's Theorem states: Let Quadrilateral ACDP be concyclic, let D,E,F respectively be the feet of the perpendiculars from P to AC, BC and AB. Then D,E,F are co-linear.

Does anyone have a proof of the converseof this theorem (i.e for three co-linear points D,E,F which lie on the feet of the perpendiculars of P with AC,BC,AB respectiveley, ABCP is concyclic), or a counterexample is no converse exists.

Thanks! LJ

Luke Johnston
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  • Look at the quadrilaterals $ADFP$ and $BEPF$, they have those right angles between one of their diagonals and one of their sides, from the perpendiculars. So, they are cyclic. Now use that the sum of their opposite angles are $180$ to solve for $\angle DFP$ and $\angle PFB$. Use that the sum of these two is $180$ for being $DFE$ a straight line. Use that to show that $\angle PAD+\angle BEP=180$. Therefore, $ACBP$ is cyclic. – OscarRascal Jan 29 '20 at 20:35

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