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Simson line

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In geometry, given a triangle ABC and a point P on its circumcircle, the three closest points to P on lines AB, AC, and BC are collinear.

Proof

In the shown diagram, we draw additional lines $AP$ and $BP$ . Then, we have cyclic quadrilaterals $ACBP$ , $PC_1A_1B$ , and $PB_1AC_1$ . (more will be added)

Simson line (main)

Simson line.png

Let a triangle $\triangle ABC$ and a point $P$ be given. Let $D, E,$ and $F$ be the foots of the perpendiculars dropped from P to lines AB, AC, and BC, respectively.

Then points $D, E,$ and $F$ are collinear iff the point $P$ lies on circumcircle of $\triangle ABC.$

Proof

Let the point $P$ be on the circumcircle of $\triangle ABC.$

$\angle BFP = \angle BDP = 90^\circ \implies$

$BPDF$ is cyclic $\implies \angle PDF = 180^\circ – \angle CBP.$

$\angle ADP = \angle AEP = 90^\circ \implies$

$AEPD$ is cyclic $\implies \angle PDE = \angle PAE.$

$ACBP$ is cyclic $\implies \angle PBC = \angle PAE \implies \angle PDF + \angle PDE = 180^\circ$

$\implies D, E,$ and $F$ are collinear as desired.

Proof

Let the points $D, E,$ and $F$ be collinear.

Simson line inverse.png

$AEPD$ is cyclic $\implies \angle APE = \angle ADE, \angle APE = \angle BAC.$

$BFDP$ is cyclic $\implies \angle BPF = \angle BDF, \angle DPF = \angle ABC.$

$\angle ADE = \angle BDF \implies \angle BPA = \angle EPF$

$= \angle BAC + \angle ABC = 180^\circ – \angle ACB \implies$

$ACBP$ is cyclis as desired.

vladimir.shelomovskii@gmail.com, vvsss

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