TITLE: Distance from one vertex of a rectangular prism to the plane determined by three other vertices, without using vectors or calculus
QUESTION [5 upvotes]: The objective is to find the shortest distance from the point $H$ to the plane $BDE$. The prism $ABCD.EFGH$ has $AB=AD=5\sqrt{2}$ and $AE=12$. I think that these numbers are badly selected by the author.

The following shows my steps to solution but I feel it is too verbose, tedious, and time consuming.
It can be solved easily with either vector (dot and cross products) or calculus (minimizing a distance function of two variables), but because this topic is for students who have not learned those subjects, I insist on only using not more than Pythagorean theorem and basic trigonometry. Analytic approach is not allowed!

Finding $DE$, $EP$ and $HP$

\begin{align*}
DE^2
&=DH^2+EH^2\\
&=12^2 + (5\sqrt2)^2 \\
&=194\\
DE
&=\sqrt{194}
\end{align*}
\begin{align*}
\frac{1}{HP^2}
&=\frac{1}{DH^2}+\frac{1}{EH^2}\\
&=\frac{1}{122^2}+\frac{1}{(5\sqrt2)^2}\\
&=\frac{97}{3600}\\
HP
&=\frac{60}{\sqrt{97}}
\end{align*}
\begin{align*}
DE \times EP &=EH^2\\
EP\sqrt{194}  &= (5\sqrt2)^2 \\
EP &=\frac{50}{\sqrt{194}}
\end{align*}
Finding $BD$, $\cos E$,  $EQ$ and $PQ$

\begin{align*}
BD 
&= \sqrt{CD^2+BC^2}\\
&= \sqrt{(5\sqrt2)^2+(5\sqrt2)^2}\\
&= \sqrt{50+50}\\
&= 10
\end{align*}
\begin{gather*}
BE^2+DE^2-2\times BE\times DE \cos E = BD^2 \\
194 + 194 - 2\times 194 \cos E = 100\\
\cos E = \frac{72}{97}
\end{gather*}
\begin{align*}
EQ 
&= EP \sec E\\
&= \frac{50}{\sqrt{194}}\times \frac{97}{72} \\
&= \frac{2425}{36\sqrt{194}}
\end{align*}
\begin{align*}
PQ 
&= \sqrt{EQ^2-EP^2}\\
&= \sqrt{\left(\frac{2425}{36\sqrt{194}}\right)^2-\left(\frac{50}{\sqrt{194}}\right)^2}\\
&= \frac{1625}{36\sqrt{194}}
\end{align*}
Finding $HQ$

\begin{align*}
HQ
&= \sqrt{EQ^2+EH^2}\\
&= \sqrt{\left(\frac{2425}{36\sqrt{194}}\right)^2 +\left(5\sqrt{2}\right)^2}\\
&= \frac{5\sqrt{15218}}{72}
\end{align*}
Finding the altitude, of $\triangle HPQ$, passing through $P$ and finding $HH'$

As the badly chosen numbers make the calculation a bit complicated. The process is left behind as your excercise. The altitude is $t = 1500/\sqrt{738073}$.
$HH'$ then can be found by equating the area from two different bases.
\begin{align*}
PQ \times HH' &= HQ \times  t\\
\frac{1625}{36\sqrt{194}} \times HH' &= \frac{5\sqrt{15218}}{72} \times \frac{1500}{\sqrt{738073}}\\
HH' &= \frac{60}{13}
\end{align*}
Thus the shortest distance from the point $H$ to the plane $BDE$ is $\tfrac{60}{13}$.

Question
Is there any shorter way to solve it but with neither using vector nor using calculus?

REPLY [3 votes]: It's worth noting that $H$ and $A$ are equidistant from the plane of $\triangle BDE$. (Consider the symmetry of the points' positions relative to the midpoint of $\overline{DE}$ and then to that plane.) If one can convince oneself of that first, then it might be easier to think-through the problem. But if not, that's okay, too.

Consider tetrahedron $HBDE$. Relative to base $\triangle HDE$, it has height $|AB|$; relative to base $\triangle BDE$, it has height $|HH'|$ (our target length). Expressing volume in two ways, we have
$$\frac13 |AB|\cdot|\triangle HDE| =  \frac13|HH'|\cdot|\triangle BDE| \tag{1}$$
Therefore, writing $M$ for the midpoint of $\overline{BD}$,
$$|HH'|=\frac{|AB|\cdot|\triangle HDE|}{|\triangle BDE|} = \frac{|AB|\cdot\tfrac12|HD||HE|}{\tfrac12|BD||ME|} = \frac{|AB||AD||AE|}{|BD||ME|} = \frac{|AB|^2|AE|}{|BD||ME|}\tag{2}$$
(Note that you'd get the same formula for $|AA'|$, with $A'$ the projection of $A$ onto the plane, by considering tetrahedron $ABDE$, which has the same volume as $HBDE$.) By Pythagoras, we have
$$\begin{align}
|BD|^2 &= |AB|^2+|AD|^2 \\[4pt]
&= 2|AB|^2 \tag{3}\\[4pt]
|ME|^2 &= |BE|^2-\left(\tfrac12|BD|\right)^2 \\
&= |AB|^2+|AE|^2-\tfrac12|AB|^2 \\
&= \tfrac12|AB|^2+|AE|^2 \tag{4}
\end{align}$$
Substituting the values $|AB|=|AD|=5\sqrt2$ and $|AE|=12$, we find
$$|BD|^2 = 100 \quad\to\quad |BD|=10 \qquad\qquad |ME|^2 = 169\quad\to\quad |ME|=13 \tag{5}$$
(These make me think that the problem's given lengths are not so "badly selected".) Therefore,

$$|HH'| = \frac{600}{130}=\frac{60}{13}\tag{$\star$}$$

which agrees with the value derived by OP. $\square$

We could have calculated $|\triangle BDE|$ via Heron's formula. Alternatively (but equivalently), we could invoke de Gua's Theorem, the dimensionally-enhanced Pythagorean Theorem for right-corner tetrahedra that deserves to be better-known:
$$|\triangle BDE|^2 = |\triangle ABD|^2 + |\triangle BAE|^2 + |\triangle BDA|^2 \tag{6}$$
so that
$$\begin{align}
|\triangle BDE|^2 &=\left(\tfrac12|AB||AD|\right)+\left(\tfrac12|AB||AE|\right)^2+\left(\tfrac12|AB||AD|\right)^2 \tag{7}\\[4pt]
&=\tfrac14\left(|AB|^2|AD|^2+|AB|^2|AE|^2+|AB|^2|AD|^2\right) \tag{8} \\[4pt]
&=\tfrac14|AB|^2|AD|^2|AE|^2\left(\frac{1}{|AB|^2}+\frac{1}{|AD|^2}+\frac{1}{|AE|^2}\right) \tag{9}
\end{align}$$
and thus, from $(2)$,

$$|HH'| = \frac{1}{\sqrt{\dfrac{1}{|AB|^2}+\dfrac{1}{|AD|^2}+\dfrac{1}{|AE|^2}}}
\tag{$\star\star$}$$

which gives the same value of $60/13$.

An analytical addendum: With $A$ at the origin, and $B$, $D$, $E$ along the $x$-, $y$-, $z$-axes, the plane of $\triangle BDE$ has intercept-intercept-intercept form
$$\frac{x}{|AB|}+\frac{y}{|AD|}+\frac{z}{|AE|}=1 \tag{10}$$
Thus, the distance from the plane to $(x,y,z)$ is given by
$$\frac{\left|\dfrac{x}{|AB|}+\dfrac{y}{|AD|}+\dfrac{z}{|AE|}-1\right|}{\sqrt{\dfrac{1}{|AB|^2}+\dfrac{1}{|AD|^2}+\dfrac{1}{|AE|^2}}} \tag{11}$$
Substituting $(x,y,z)\to A=(0,0,0)$ or $(x,y,z)\to H=(0,|AD|,|AE|)$ gives $(\star\star)$.