$ABCD$ is a parallelogram, $Q$ lies on $AC$ such that $AQ:QC=1:3$. $DQ$ intersects $AB$ and $BC$ in $M$ and $P$, respectively. I should find $AM:MB$, $BC:PC$ and $DM:QP$.
We have NOT studied similar triangles, and I should solve the problem by using Thales's theorem. Which angle should we look at for the first ratio? What about $\angle AQD$?
I guess in this problem $\triangle AMQ \sim \triangle DQC$, but I cannot use this.
Best Answer
Thales theorem tells us that
$${DC\over AM}={QC\over AQ}=3$$
Because $CD=AB=AM+MB$ we have
$${MB\over AM}=2$$