Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane

From ProofWiki
Jump to navigation Jump to search

Theorem

In the words of Euclid:

If a straight line be set up at right angles to two straight lines which cut one another, at their common point of section, it will also be at right angles to the plane through them.

(The Elements: Book $\text{XI}$: Proposition $4$)


Proof

Euclid-XI-4.png

Let $AB$ and $CD$ be two straight lines which cut one another at the point $E$.

Let a straight line $EF$ be set up at right angles to $AB$ and $CD$.

It is to be demonstrated that $EF$ is also at right angles to the plane through $AB$ and $CD$.


Let $AE$, $EB$, $CE$ and $ED$ be cut off equal to one another.

Let $AD$ and $BC$ be drawn.

Let a straight line $GEH$ be drawn arbitrarily across from $AD$ and $BC$ through $E$ such that $G$ lies on $AD$ and $H$ lies on $BC$.

Let $FA$, $FG$, $FD$, $FC$, $FH$ and $FB$ be joined from the point $F$.

We have that $AE$ and $ED$ are equal to the two straight lines $CE$ and $EB$.

From Proposition $15$ of Book $\text{I} $: Two Straight Lines make Equal Opposite Angles:

$\angle AED = \angle CEB$

Therefore the base $AD$ of $\triangle AED$ equals the base $BC$ of $\triangle CEB$.

Therefore by Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$\angle DAE = \angle EBC$

But from Proposition $15$ of Book $\text{I} $: Two Straight Lines make Equal Opposite Angles:

$\angle AEG = \angle BEH$

Therefore $\triangle AGE$ and $\triangle BEH$ are two triangles which have:

two angles equal to two angles respectively
one side $AE$ equal to one side $EB$, that adjacent to the equal angles.

Therefore from Proposition $26$ of Book $\text{I} $: Triangle Angle-Side-Angle and Side-Angle-Angle Congruence:

the remaining sides of $\triangle AGE$ and $\triangle BEH$ will also be equal.

Therefore $GE = EH$ and $AG = BH$

We have that:

$AE = EB$

while $FE$ is common and at right angles.

Therefore by Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$FA = FB$

For the same reason:

$FC = FD$

We have that:

$AD = CB$

and:

$FA = FB$

Therefore $FA = FB$ and $AD = BC$

But $FD$ is proved equal to $FC$.

Therefore by Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Congruence:

$\angle FAD = \angle FBC$

Similarly:

$AG = BH$

and:

$FA = FB$

and so $FA = AG$ and $FB = BH$.

We also have that:

$\angle FAG = \angle FBH$

and so from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$FG = FH$


We have that $GE = EH$.

We also have that $EF$ is common.

So the two sides $GE$ and $EF$ are equal to the two sides $HE$ and $EF$.

The side $FG$ equals the side $FH$.

Therefore from Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Congruence:

$\angle GEF = \angle HEF$

Therefore each of $\angle GEF$ and $\angle HEF$ is right.

Therefore $FE$ is at right angles to $GH$ drawn at random through $E$.

Similarly it can be proved that $FE$ will make right angles with all the straight lines which meet it and are in the plane of reference.

So from Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:

$FE$ is at right angles to the plane of reference.

But the plane of reference is the plane through the straight lines $AB$ and $CD$.

Hence the result.

$\blacksquare$


Historical Note

This proof is Proposition $4$ of Book $\text{XI}$ of Euclid's The Elements.


Sources