A cube $ABCD A_1 B_1 C_1 D_1$ is given, with edge $a$. On the edge $C_!D_1$we take a point $L$, with $C_1L=\dfrac{3a}{4}$; on the edge $A_1B_1$ we take point $M$, with $A_1M=\dfrac{a}{2}$; on the edge $BB_1$, point $N$ is taken, with $B_1N=\dfrac{a}{4}$. Calculate the perimeter of the section determined in the cube, by a plane guided by points $L$, $M$ and $N$.
The answer I got is correct but I considered that the triangles $MB_1N$ and $LC_!E$ are similar but I was unable to demonstrate it. Would it be possible to demonstrate this similarity or is there another way to resolve it?
$\triangle MNB1:MN^2=(\frac{a}{4})^2+(\frac{a}{2})^2 \implies \boxed{MN = \frac{a\sqrt5}{4}}\\ MO \perp D_1C_1 (O \in D_1C_1)\implies \triangle MLO: LM^2 = a^2+(\frac{a}{4})^2 \implies \boxed{LM=\frac{a\sqrt{17}}{4}}\\ \triangle MB_1N \sim \triangle LC_1E :\\ C1E.\frac{a}{2}=\frac{a}{4}.\frac{3a}{4} \therefore C_1E = \frac{3a}{8}\\ \triangle LC_1E: LE^2 = (\frac{3a}{4})^2+(\frac{3a}{8})^2 \therefore \boxed{LE =\frac{3a\sqrt5}{8} }\\ P \perp C_1E:(P \in C_1E) \implies \triangle NPE: NE^2 = a^2+(\frac{3a}{8}-\frac{a}{4})^2 \therefore \boxed{NE=\frac{a\sqrt65}{8}}\\ 2p =\frac{a\sqrt5}{4} +\frac{3a\sqrt5}{8}+\frac{a\sqrt{17}}{4}+\frac{a\sqrt{65}}{8}=\\ \frac{5a\sqrt5}{8}+\frac{a\sqrt{17}}{4}+\frac{a\sqrt{65}}{8}=\boxed{\frac{a}{8}(5\sqrt5+2\sqrt{17}+\sqrt{65})}$
Hint: If you have to show that triangles $LC_1E$ and $MB_1N$ are similar, project $LN$ on the plane having triangle $LC_1E$, call it $M'N'$, now in this plane, as I explained in comment we have:
$MN||LE\Rightarrow LE||M'N'$
which deduces two triangle are similar.