## Πέμπτη, 23 Φεβρουαρίου 2012

### PP*

Let ABC be a triangle, P a point and A1B1C1 the pedal triangle of P.

Let Q be a point on the line PDP* [D = the center of the common pedal circle of P and its isog. conjugate P*] and Q1Q2Q3 the circumcevian triangle of Q with respect A1B1C1.

The triangles ABC, Q1Q2Q3 are perspective.

APH 23 February 2012

#### 1 σχόλιο:

1. Lemma:-
Given a triangle ABC, A1B1C1 and A2B2C2 are the circumcevian triangles of two points P,Q wrt ABC. Then the triangle formed by A1A2,B1B2,C1C2 is perspective to ABC.

Proof:-
If A1A2 intersects BC at A3, then BA3/CA3=BA1.BA2/CA1.CA2 and similarly for B3,C3. By the trigonometric form of Ceva's theorem, BA1/CA1.CB1/AB1.AC1/BC1=1 and similar for A2,B2,C2. So BA3/CA3.CB3/AB3.AC3/BC3=1. So by Menelaus' theorem A3,B3,C3 are collinear. So by Desergaus theorem, the triangle formed by A1A2,B1B2,C1C2 is perspective to ABC.

In the given problem, if A2B2C2 is the pedal triangle of P*, then from your previous problem we get that A2B2C2 and Q1Q2Q3 are perspective and we also have A1B1C1 and Q1Q2Q3 are perspective. So using the lemma we get that ABC and Q1Q2Q3 are perspective.