Let $ ABCD$ be a quadrilateral, and $ E$ be intersection points of $ AB,CD$ and $ AD,BC$ respectively. External bisectors of $ DAB$ and $ DCB$ intersect at $ P$, external bisectors of $ ABC$ and $ ADC$ intersect at $ Q$ and external bisectors of $ AED$ and $ AFB$ intersect at $ R$. Prove that $ P,Q,R$ are collinear.
Problem
Source: Iranian National Olympiad (3rd Round) 2008
Tags: geometry, projective geometry, geometry proposed
darij grinberg
12.09.2008 14:51
Should be one of the results in http://www.mathlinks.ro/viewtopic.php?t=30921 . darij
sumita
12.09.2008 14:58
$ AD \cap BC = F,AB \cap DC = E$ external bisector $ DAB = L_{1}$ external bisector $ ABC = L_{2}$ external bisector $ BCD = L_{3}$ external bisector $ CDA = L_{4}$ external bisector $ AFB = L_{5}$ external bisector $ AED = L_{6}$ $ L_{3} \cap L_{4} = A_{1}$ $ L_{3} \cap L_{5} = C_{1}$ $ L_{5} \cap L_{4} = B_{1}$ $ L_{1} \cap L_{2} = A_{2}$ $ L_{2} \cap L_{6} = B_{2}$ $ L_{1} \cap L_{6} = C_{2}$ but: $ A_{1}B_{1} \cap A_{2}B_{2} = Q$ $ A_{1}C_{1} \cap A_{2}C_{2} = P$ $ C_{1}B_{1} \cap C_{2}B_{2} = R$ but$ A_{1}A_{2},B_{1}B_{2},C_{1}C_{2}$ go through incircle of $ DFC$ now from desargue theorem we coclude $ Q,P,R$ are collinear.
Sx763_
02.03.2016 00:37
sumita wrote: $ AD \cap BC = F,AB \cap DC = E$ external bisector $ DAB = L_{1}$ external bisector $ ABC = L_{2}$ external bisector $ BCD = L_{3}$ external bisector $ CDA = L_{4}$ external bisector $ AFB = L_{5}$ external bisector $ AED = L_{6}$ $ L_{3} \cap L_{4} = A_{1}$ $ L_{3} \cap L_{5} = C_{1}$ $ L_{5} \cap L_{4} = B_{1}$ $ L_{1} \cap L_{2} = A_{2}$ $ L_{2} \cap L_{6} = B_{2}$ $ L_{1} \cap L_{6} = C_{2}$ but: $ A_{1}B_{1} \cap A_{2}B_{2} = Q$ $ A_{1}C_{1} \cap A_{2}C_{2} = P$ $ C_{1}B_{1} \cap C_{2}B_{2} = R$ but$ A_{1}A_{2},B_{1}B_{2},C_{1}C_{2}$ go through incircle of $ DFC$ now from desargue theorem we coclude $ Q,P,R$ are collinear. go through excircle od $ DFC $