$ABCD$ is a parallelogram in which $\frac{AB}{BC} = \lambda $. If $P$, $Q$ are points on the line $CD$ with $P$ on $CD$ and $M$ is a point $BC$ such that $AD$ bisects $\angle PAQ$ and $AM$ bisects $\angle PAB$, $BM = a$, $DQ = b$, prove that $AQ = \frac{a}{\lambda + b} $.
I have tried to use the fact that $\frac{AB}{BC} = \lambda$ by constructing the internal angle bisector of $\angle B$ meeting $AC$ at $O$, but could not proceed much further after that.
This is NOT an answer but a query instead. I try to draw a special case of it by letting (1) $\lambda = 1$; and (2) AB = AP = AM.
Clearly, $a \lt 1$ and $\dfrac {a}{\lambda + b} = \dfrac {a}{1 + b}$is a quantity less than 1 but the picture shows AQ is a line much longer than 1. Am I missing something?