Problem 2. Let $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be a cube with side $A B=a$. Consider points $E \in(A B)$ and $F \in(B C)$ such that $A E+C F=E F$.
a) Determine the measure of the angle formed by the planes $\left(D^{\prime} D E\right)$ and $\left(D^{\prime} D F\right)$.

b) Calculate the distance from $D^{\prime}$ to the line $E F$.
Solution. a) We extend the segment $[BA]$ with the segment $[AH] \equiv [FC]$. Then $\triangle DAH \equiv \triangle DCF$ (S.S.), from which $\widehat{HDA} \equiv \widehat{FDC}$, so $m(\widehat{HDF}) = 90^{\circ}$.

Then, $[DH] \equiv [DF]$, from which $\triangle DHE \equiv \triangle DFE$ (S.S.S.) and from here $m(\widehat{FDE}) = m(\widehat{HDE}) = 45^{\circ}$ $3 p$

![](https://cdn.mathpix.com/cropped/2024_06_07_86781efbf6608150f54eg-3.jpg?height=65&width=1304&top_left_y=859&top_left_x=432)

![](https://cdn.mathpix.com/cropped/2024_06_07_86781efbf6608150f54eg-3.jpg?height=62&width=1369&top_left_y=912&top_left_x=367)

b) We denote by $P$ the projection of point $D$ onto the line $EF$. According to the theorem of the three perpendiculars, we get $D'P \perp EF$, so $d(D', EF) = D'P \ldots \mathbf{1 p}$

From the congruence $\triangle DHE \equiv \triangle DFE$, we get $DP = AD = a \ldots \ldots \ldots 1 \mathbf{p}$

The Pythagorean theorem leads to $D'P = a \sqrt{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots 1 \mathbf{1 p}$