Ratio between subsegments of space diagonal of a cube

Question

Let $ABCDEFGH$ be a cube where $ABCD$ is the ground face and $E$ is about $A$, $F$ above $B$, and so on. Consider the space diagonal $AG$, and call $P$ the orthogonal projection of $B$ on $AG$.

The question is to show that $2|AP| = |PG|$. One can easily calculate that the space diagonal has length $\sqrt{3}$, and by the congruence of $ABG$ and $ABP$ we find that $|AP| = \sqrt{3}/3$. Hence $|PG| = 2\sqrt{3}/3$, and we are done. This, however, requires some calculations.

Can we show $2|AP| = |PG|$, without calculating any of the lengths? So, by some similarity or symmetry argument. One observation is that $P$ is in the plane $BDE$, so by symmetry there is a point $P'$ in the plane $CFH$, and in this setup it is sufficient to prove that $|AP| = |PP'|$.

Answer 1

Construction: Join $BG$.

Let the lenght of side $AB$ be $a$, so $AG=a\sqrt{3}$. By Pythagoras' theorem, $BG=a\sqrt{2}$.

Clearly, $BG$ is perpendicular to $AB$ as the whole plane $BCGF$ is perpendicular to $AB$. As a result, we get a right angled $ΔABG$ where $BP⊥AG$.

Clearly, $ΔAPB \sim ΔABG$, so$$\frac{AP}{AB}=\frac{AB}{AG} \Longrightarrow \frac{AP}{a}=\frac{a}{a\sqrt{3}} \Longrightarrow AP=\frac{a}{\sqrt{3}}.$$

Similarly, $ΔBPG \sim ΔABG$, so$$\frac{PG}{BG}=\frac{BG}{AG} \Longrightarrow \frac{PG}{a\sqrt{2}}=\frac{a\sqrt{2}}{a\sqrt{3}} \Longrightarrow PG=\frac{2a}{\sqrt{3}}.$$

Now, $$\frac{AP}{PG}=\frac{1}{2},$$

therefore $$2|AP|=|PG|.$$

Quod erat demonstrandum…