the question
Consider the regular quadrilateral pyramid $VABCD$ with the vertex in $V$ and the points $M, N, P$ the means of the segments $AD, BC, VA$. Show that the angle of the line $CP$ with the plane $(BAD)$ has a measure of $45$ if and only if the angle of the line $CP$ with the plane $(VMN)$ has a measure of $30$.
The idea
The drawing
First, let's clarify the angles between the lines and the planes.
$1)$ $\angle(CP,(BAD))$
Let $PX \perp (BAD)$ and we know that $C\in(BAD)=> \angle(CP,(BAD))=\angle PCX$
Also, we can demonstrate that X is the midpoint of AO.
$2)$ $\angle(CP,(VMN))$
Let $PY \perp (VMN)$ and we know that $CN \perp (VMN)=> \angle(CP,(VMN))= \angle (CP,YN)$
We can simply show that Y is the midpoint of line VM
$YP||CN$ and $CN>PY$ so that means that $PYCN$ is a trapeze.
Let $BC=4l$
From now we have two cases:
case_1: We know that $\angle PCX=45$. we show that $\angle (CP,YN)=30$
case_2: We show that $\angle PCX=45$. we know that $\angle (CP,YN)=30$
case_1:
$\angle PCX=45$ means that traingle $PXC$ is right angled and issoscels $=> PX=XC=3l\sqrt{2}=> PC=6l$
Let $YZ \perp (ABCD)$ and $ZN=3l, PX=YZ=3l\sqrt{2}$
From here we get that $YN$ doesn't equal $PC$ so the trapezoid PYCN isn't isosceles.
I don't know what to do forward. I hope one fo you can help me!