The question
The regular quadrilateral pyramid $VABCD$ has the vertex $V$. Let $M$ be the middle of the edge $AD$ and $E$ be the point of intersection of the lines $AC$ and $BM$. Prove that the triangles $VAC, EAV$ are similar if and only if the angle formed by the right $VE$ with the plane $(VBD)$ measures $30°$.
The drawing
the angle formed by the right $VE$ with the plane $(VBD)$ its actually angle $EVO=30°$
let $AC=6x => AE=2x, EC=4x, EO=x, AO=3x$
Case 1: We know that $\angle EVO=30$. We show that $VAC$ and $EAV$ are similar.
$\angle EVO=30°, \angle VOE=90°=> VE=2x => VO=x\sqrt{3} => VA=2x\sqrt{3}$
From here we can say that its verified $\frac{VA}{AE}=\frac{AC}{AV}=\frac{VC}{VE}=\sqrt{3}$ which makes $VAC, EAV$ similar.
Case 2: We show that $\angle EVO=30°$. We know that $VAC$ and $EAV$ are similar.
From the similarity we get that $\frac{VA}{AE}=\frac{AC}{AV}=\frac{VC}{VE}=> VA^2=AC\times AE=12x^2=> VA=2x\sqrt{3},AO=3x=> VO=x\sqrt{3}=> \angle VAO=30°=> \angle AVO=60$
we can show that $\frac{AE}{EO}=\frac{AV}{VO}=2$ which makes $VE$ an bisector $=> \angle EVO=30°$
I'm not sure if my calculation and rationament (the bisector thing) are right. Hope one of you can help me! Thank you!
Case 1: Well done; you could have just specified that $VC=VA(=2x\sqrt3)$
Case 2: Well done; you could have quoted the angle bissector theorem and perhaps write more classicaly $\frac{V\color{red}A}{V\color{red}O}=\frac{\color{red}AE}{\color{red}OE}$ :)