# 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