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**Proof of the Simpson
Line**

**By: Ginger Rhodes**

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**Given:** DABC is inscribed in
circle O, The Pedal point P is on the circle O, PS^AC, PT^AB, and PR^BC [These are segments]

**Prove:** Points R, S, and T, the Simpson line, are
collinear

Proof:

<PSA and <PTA are both right
angles, and therefore supplementary. So <SPT and <TAS are supplementary
and opposite angles in Quadrilateral PSAT are supplementary.

Now,
<PST
and <PAT
have the same intercepted arc, which means the angles are congruent. Similary,
points P, R, C, and S are cyclic.

So,
<PCR
and <PSR
have the same intercepted arc, which means the angles are congruent.

By
hiding some of the lines it is easier to see DPAB@DPCB [same intercepted
arcs].

So, <PST congruent
<PAT, <PCR congruent
<PSR, and <PAB congruent <PCB.

Notice, <PAB is
another name for <PAT and <PCB is another name for <PCR.

Using the transitive property we can establish <PST @ <PSR, which
implies S, T, and R are collinear!