Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane

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 :
 * $\angle AED = \angle CEB$

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

Therefore by :
 * $\angle DAE = \angle EBC$

But from :
 * $\angle AEG = \angle BEH$

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 :
 * 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 :
 * $FA = FB$

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 :
 * $\angle FAD = \angle FBC$

Similarly:
 * $AG = BH$

and:
 * $FA = FB$

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

We also have that:
 * $\angle FAG = \angle FBH$

and so from :
 * $FG = FH$

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 :
 * $\angle GEF = \angle HEF$

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 :
 * $FE$ 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.