## Solution:

a) From the Pythagorean theorem, we obtain: $A M=4 \sqrt{2}, A C=8 \sqrt{2}$ and $M C=4 \sqrt{6}$, and from the converse of the Pythagorean theorem, we obtain that triangle $A M C$ is right-angled at $M$
b) Let $(A M C) \cap(E F H)=d$. We get $d\|A C\| E G$ and since $E M \| G N$, it follows that
MNGE is a parallelogram. $\mathrm{FN}=12 \mathrm{~cm}$ ..... 2p
c) Let $M P \| E A$ and since $E A \perp(A B D)$, we obtain $M P \perp(A B D)$ and $M P=4 \mathrm{~cm}$ ..... $1 p$
If $P Q \perp A C$, it follows that the angle corresponding to the dihedral angle is $M Q P$.
The tangent of the angle between the planes $(A B D)$ and $(A C M)$ has the value $\sqrt{2}$ ..... $1 p$
## SUBJECT IV

$A B C D E F G H$ is a regular quadrilateral prism with $A B=8 \mathrm{~cm}$ and $A E=4 \mathrm{~cm}$, and $M$ is the midpoint of the edge $(E H)$.

a) Prove that the line $A M$ is perpendicular to the line $M C$.

b) Find the length of the segment $(F N)$ knowing that $(A C M) \cap F G=\{N\}$.

c) Find the tangent of the angle between the planes $(A B D)$ and $(A C M)$.