2024 INMO
==Problem 1
\text {In} triangle ABC with 
, \text{point E lies on the circumcircle of} \text{triangle ABC such that} 
. \text{The line through E parallel to CB intersect CA in F} \text{and AB in G}.\text{Prove that}\\ \text{the centre of the circumcircle of} triangle EGB \text{lies on the circumcircle of triangle ECF.}
Solution 1
\includegraphics[width=1.25\textwidth]{INMO 2024 P1.png} To Prove: Points E,F,P,C are concyclic \newpage
Observe: ![\[\angle CAB=\angle CBA=\angle EGA\]](http://latex.artofproblemsolving.com/8/0/4/804eea72b35117cbeb648bd29eb74cd965fc3103.png)
![\[\angle ECB=\angle CEG=\angle EAB= 90^\circ\]](http://latex.artofproblemsolving.com/7/2/2/722712b65d8613d3797ea63c4f56a91c07bc74bb.png)
\text{Notice that} ![\[\angle CBA = \angle FGA\]](http://latex.artofproblemsolving.com/d/2/a/d2a5044184909fe92d8505f40511b9321c91a798.png)
because 
} \implies 
\:\text{or} \:
.\\
.\\
\text{Here F is the circumcentre of \traingle EAG becuase F lies on the Perpendicular bisector of AG.}\\\\
\implies \text{
is the midpoint of 
} \implies \text{
is the perpendicular bisector of .}\\
\text{This gives} \:![\[\angle EFP =90^\circ\]](http://latex.artofproblemsolving.com/0/8/f/08ffa237cee2e795c91719df6617a5e70836a3a0.png)
.\\
\text{And because}![\[\angle EFP+\angle ECP=180^\circ\]](http://latex.artofproblemsolving.com/7/9/e/79ec6002f0be9c7e10f9e7c7b5fc805c0bc90903.png)
. \:\text{Points E,F,P,C are concyclic.}\\
\text{Hence proven that the centre of the circumcircle of 
lies on the circumcircle of 
.}
