## Task 6 - 210936

For a tetrahedron $A B C D$, let the edge lengths be $A B=10 \mathrm{~cm}, B C=6 \mathrm{~cm}, A C=8 \mathrm{~cm}, A D=13$ $\mathrm{cm}, B D=13 \mathrm{~cm}$, and the perpendicular from $D$ to the plane of triangle $A B C$ be $12 \mathrm{~cm}$ long.

Prove that these conditions uniquely determine the length of the edge $C D$, and find this edge length!
Let $L$ be the foot of the perpendicular from $D$ to the plane of triangle $\triangle A B C$. Then the triangles $\triangle A D L$ and $\triangle B D L$ are right-angled at $L$ and it holds that

$$
|A L|=\sqrt{|A D|^{2}-|D L|^{2}}=\sqrt{13^{2}-12^{2}} \text{ cm}=\sqrt{25} \text{ cm}=5 \text{ cm}
$$

and analogously $|B L|=5 \text{ cm}$. Since $|A B|=10 \text{ cm}$ is the sum of these two segments, $L$ lies on the segment $A B$ and forms its midpoint.

Since $|A B|^{2}=|B C|^{2}+|A C|^{2}$, triangle $\triangle A B C$ is right-angled at $C$. Therefore, by the converse of Thales' theorem, $|A L|=|B L|=|C L|=5 \text{ cm}$.

Triangle $\triangle C D L$ is right-angled at $L$. Thus, by the Pythagorean theorem,

$$
|C D|=\sqrt{|C L|^{2}+\left|D L^{2}\right|}=\sqrt{5^{2}+12^{2}} \text{ cm}=13 \text{ cm}
$$