Let $ABC$ a triangle with $AB=BC$ and incircle $\omega$. Let $M$ the mindpoint of $BC$; $P, Q$ points in the sides $AB, AC$ such that $PQ\parallel BC$, $PQ$ is tangent to $\omega$ and $\angle CQM=\angle PQM$. Find the perimeter of triangle $ABC$ knowing that $AQ=1$.

Initially, it can be seen that PQ//BC, take the points of tangency K ∈ BP, L ∈ PQ and N ∈ QC.PK=PL,QL=QN. Take segment AP as X,AK+AN=2X+1,AN=CN=(2X+1)/2 and QC=2X,by similarity X/1=BC/2X+1, then BC=2X²+X,BP=2X²,by parallelism,MC=QC,so MC=2X and BM=2X,equating the information, we have 2X²+X=4X,then X=3/2. being the perimeter of the triangle 4X²+4X+1,the perimeter is 16