We have a parallelogram $ABCD$, $Q$ lies on $BC$ such that $BQ:QC=2:3$. The line $AQ$ intersects the diagonal $DB$ and the line $DC$ in $M$ and $P$, respectively. I should find $DM:MB$, $DC:PC$ and $AM:QP$. (I have not studied similar triangles and should solve the problem by using Intercept theorem as I started.)
$AQ \cap BD=M, AD \parallel BC$ and $AD \cap AQ;BD=A,D$ and $BC \cap AQ;BD=Q,B$. Intercept theorem tells us that $\dfrac{AD}{BQ}=\dfrac{AM}{MQ}=\dfrac{DM}{MB}$. But $AD=BC$, therefore $\dfrac{BC}{BQ}=\dfrac{AM}{MQ}=\dfrac{DM}{MB}=\dfrac{5}{2}$. Similarly, $\dfrac{CP}{PD}=\dfrac{CQ}{AD}=\dfrac{PQ}{PA}$ $(AP\cap DP=P; AD \parallel BC)$. But $AD=BC$, therefore $\dfrac{CP}{DP}=\dfrac{CQ}{BC}=\dfrac{PQ}{PA}=\dfrac{3}{5}$ and $\dfrac{DC}{PC}=\dfrac{2}{3}$. How can I find $AM:QP$ by using Intercept theorem? I am so messed up and I will be very grateful if you could tell me some tricks to use in such situations. How to approach this kind of problems in general?
You already know that ${AM\over MQ}={5\over2}$, that is: ${AM\over AQ}={5\over7}$. Hence: $$ {AM\over QP}={AM\over AQ}{AQ\over QP}={5\over7}\cdot{2\over3}={10\over21}. $$