Let $ABCDEFGH$ be a cube of side length 5, as shown. Let $P$ and $Q$ be points on $\overline{AB}$ and $\overline{AE}$, respectively, such that $AP = 2$ and $AQ = 1$. The plane through $C$, $P$, and $Q$ intersects $\overline{DH}$ at $R$. Find $DR$.
Here's my thoughts so far. Since triangles APQ and DPR are similar, we can use the proportion $\frac{2}{1}=\frac{7}{DR}$, therefore DR=3.5. Is this correct? Thanks! Please explain if not correct.
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Let we put a reference system with the origin at $A$, the $x$-axis along $AB$, the $y$-axis along $AD$ and the $z$-axis along $AE$. Then we have a plane $\pi$ through $$ P(2;0;0),\quad C(5;5;0),\quad Q(0;0;1) $$ with equation $$ 5x-3y+10z = 10 $$ hence if $x=0$ and $y=5$ (corresponding to $D$) we have $z=\color{red}{\large\frac{5}{2}}$ (corresponding to $DR$).
You are wrong. The proper similar triangles are $APQ$ and $DCR$.
Simply extend the segments $CP$ and $RQ$ until they intersect at a common point which is the common intersection point $S$ of the planes $ABCD, \, ADHE$ and $PQRC$ which is on the line $AD$. In other words $CP, \, RQ$ and $DA$ intersect in $S$. Then you see that $APQ$ and $DCR$ are similar triangles because a homothety with center $S$ maps one to the other. And then simply $$\frac{DR}{AQ} = \frac{CD}{PA}$$ which turns into $$DR = \frac{DR}{1} = \frac{5}{2} = 2.5$$
Actually, it is probably easier to observe that $AQ \parallel DR \,$, $\,\, AP \parallel CD$ and $PQ \parallel CR$, so triangles $APQ$ and $DCR$ are similar.