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**Third Pedal Triangle**

By Thuy Nguyen

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__Theorem__:
The third pedal triangle is similar to the original triangle.

Click here for the GSP script for a general construction of a pedal triangle ABC where P is any point in the plane ABC.

Click here for the GSP script to see the third pedal triangle.

Proof:

We first
note that P lies on the circumcircles of all the triangles AB_{1}C_{1},
A_{2}B_{1}C_{2}, A_{2}B_{2}C_{1},
and A_{3}B_{2}C_{3}. Now join P to every vertex on all
four triangles. Then we have

<C_{1}AP
= <C_{1}B_{1}P = <A_{2}B_{1}P = <A_{2}C_{2}P

= <B_{3}C_{2}P
= <B_{3}A_{3}P

and

<PAB_{1}
= <PC_{1}B_{1} = <PC_{1}A_{2} = <PB_{2}A_{2}

= <PB_{2}C_{3}
= <PA_{3}C_{3}.

That is,
the two pairs into which AP divides <A have their equal counterparts at B_{1}
and C_{1}, again at C_{2} and B_{2}, and finally both
at A_{3}. So ▲ABC and ▲A_{3} B_{3} C_{3}
have equal angles at A and A_{3}. Similarly, they have equal angles at
B and B_{3}. QED.