Help in solving this geometric inequality I have the following geometry problem with me:
"The altitudes through the vertices A,B,C of an acute angles triangle meet the opposite sides at D,E and F respectively, and AB > AC. The line EF meets BC at P, and the line through D parallel to EF meets the lines AC and AB at Q and R respectively. Let N be a point on BC such that $\angle{NQP}  + \angle{NRP} < 180 $. Prove that BN > CN."
My try:
Let J be the point of intersection of QR and BF. Since E,F,P,Q and R are independent of the choice of N, I try to find as many properties involving the above points. However, the only property which might be quite useful that I have deduced is that FPQJ is a parallelogram. I am unsure of whether even that is useful for this problem. Could anybody please help me with this!
 A: As I stated in my comment, the problem is equivalent to showing that $PQRM$ is cyclic, where $M$ is the midpoint of $BC$ - do you see why? (In a cyclic quadrilateral, opposite angles add up to $180^\circ$ - this was motivated by the condition $\angle NQP + \angle NRP <180^\circ$.)
We first show that $RBQC$ is cyclic. By definition, $\angle BFC=\angle BEC=90$ so $BFEC$ is cyclic. Then $\angle RQC=\angle PEC = \angle RBC$ hence $RBQC$ is cyclic. Once we know $RBQC$ is cyclic, we know that $RD\cdot DQ=BD\cdot DC$ by power of a point. To prove that $PQMR$ is cyclic, it is sufficient to show that $PD\cdot DM=RD\cdot DQ$ for the same reasons.
Now, by Ceva and Menelaus on $\triangle ABC$ we have that $\frac{CD}{BD}=\frac{CP}{BP}\implies CD\cdot BP= CP\cdot BD$ (computation is left to the reader). Let $PM,BM,$ and $DM$ be $x,y,$ and $z$ respectively. Then $CD=y+z$, $BP=x-y$, $CP=x+y$ and $BD=y-z$ so $$\begin{align*} CD\cdot BP&= CP\cdot BD \\  \implies (x-y)(y+z)&=(x+y)(y-z) \\ \implies xy+xz-y^2-yz&= xy-xz+y^2-yz \\ \implies xz=y^2 \end{align*} $$
Hence $PD\cdot DM =z(x-z)=xz-z^2=y^2-z^2=(y+z)(y-z)=BD\cdot DC$
So $PD\cdot DM=BD\cdot DC=RD\cdot DQ$, hence $PQMR$ is cyclic as desired!
