Common Section of Bisecting Planes of Cube Bisect and are Bisected by Diagonal of Cube
Theorem
In the words of Euclid:
- If the sides of the opposite planes of a cube be bisected, and planes be carried through the points of section the common section of the planes and the diameter of the cube bisect one another.
(The Elements: Book $\text{XI}$: Proposition $38$)
Proof
Let $AF$ be a cube.
Let the edges of the opposite faces $CF$ and $AH$ of $AF$ be bisected at the points $K, L, M, N, O, P, Q, R$.
Let the plane $KN$ be constructed through $K, L, M, N$.
Let the plane $OR$ be constructed through $O, P, Q, R$.
Let $US$ be the common section of $KN$ and $OR$.
Let $DG$ be the diameter of the cube $AF$.
It is to be demonstrated that:
- $UT = TS$
- $DT = TG$
Let $DU, UE, BS, SG$ be joined.
We have that $DO$ is parallel to $PE$.
Therefore from Proposition $29$ of Book $\text{I} $: Parallelism implies Equal Alternate Angles:
- $\angle DOU = \angle UPE$
We have that:
- $DO = PE$
- $OU = UP$
- $\angle DOU = \angle UPE$
So from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:
- $\triangle DOU = \triangle PUE$
Therefore:
- $\angle OUD = \angle PUE$
So from Proposition $14$ of Book $\text{I} $: Two Angles making Two Right Angles make Straight Line:
- $DUE$ is a straight line.
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For the same reason:
- $BSG$ is a straight line.
and:
- $BS = SG$
We have that:
- $CA = DB$ and $CA \parallel DB$
and:
- $CA = EG$ and $CA \parallel EG$
Therefore fromProposition $9$ of Book $\text{I} $: Lines Parallel to Same Line not in Same Plane are Parallel to each other:
- $DB = EG$ and $DB \parallel EG$
We have that the straight lines $DE$ and $BG$ join the endpoints of $DB$ and $EG$.
Therefore from Proposition $33$ of Book $\text{I} $: Lines Joining Equal and Parallel Straight Lines are Parallel:
- $DE \parallel BG$
Therefore from Proposition $29$ of Book $\text{I} $: Parallelism implies Equal Alternate Angles:
- $\angle EDT = \angle BGT$
Therefore from Proposition $29$ of Book $\text{I} $: Two Straight Lines make Equal Opposite Angles:
- $\angle DTU = \angle GTS$
Therefore from Proposition $26$ of Book $\text{I} $: Triangle Side-Angle-Angle Congruence:
- $\triangle DTU = \triangle GTS$
Therefore:
- $DT = TG$
and
- $UT = TS$
$\blacksquare$
Historical Note
This proof is Proposition $38$ of Book $\text{XI}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions