Simson line

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)

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.

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

$AEPD$ 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.$

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Simson line

Proof

Simson line (main)