# Proof: By Euclid

• For let $AE$, $EB$, $CE$ and $ED$ have been cut off from (the two straight lines so as to be) equal to one another.
• And let $GEH$ have been drawn, at random, through $E$ (in the plane passing through $AB$ and $CD$).
• And let $AD$ and $CB$ have been joined.
• And, furthermore, let $FA$, $FG$, $FD$, $FC$, $FH$, and $FB$ have been joined from the random (point) $F$ (on $EF$).
• For since the two (straight lines) $AE$ and $ED$ are equal to the two (straight lines) $CE$ and $EB$, and they enclose equal angles [Prop. 1.15], the base $AD$ is thus equal to the base $CB$, and triangle $AED$ will be equal to triangle $CEB$ [Prop. 1.4].
• Hence, the angle $DAE$ [is] equal to the angle $EBC$.
• And the angle $AEG$ (is) also equal to the angle $BEH$ [Prop. 1.15].
• So $AGE$ and $BEH$ are two triangles having two angles equal to two angles, respectively, and one side equal to one side - (namely), those by the equal angles, $AE$ and $EB$.
• Thus, they will also have the remaining sides equal to the remaining sides [Prop. 1.26].
• Thus, $GE$ (is) equal to $EH$, and $AG$ to $BH$.
• And since $AE$ is equal to $EB$, and $FE$ is common and at right angles, the base $FA$ is thus equal to the base $FB$ [Prop. 1.4].
• So, for the same (reasons), $FC$ is also equal to $FD$.
• And since $AD$ is equal to $CB$, and $FA$ is also equal to $FB$, the two (straight lines) $FA$ and $AD$ are equal to the two (straight lines) $FB$ and $BC$, respectively.
• And the base $FD$ was shown (to be) equal to the base $FC$.
• Thus, the angle $FAD$ is also equal to the angle $FBC$ [Prop. 1.8].
• And, again, since $AG$ was shown (to be) equal to $BH$, but $FA$ (is) also equal to $FB$, the two (straight lines) $FA$ and $AG$ are equal to the two (straight lines) $FB$ and $BH$ (respectively).
• And the angle $FAG$ was shown (to be) equal to the angle $FBH$.
• Thus, the base $FG$ is equal to the base $FH$ [Prop. 1.4].
• And, again, since $GE$ was shown (to be) equal to $EH$, and $EF$ (is) common, the two (straight lines) $GE$ and $EF$ are equal to the two (straight lines) $HE$ and $EF$ (respectively).
• And the base $FG$ (is) equal to the base $FH$.
• Thus, the angle $GEF$ is equal to the angle $HEF$ [Prop. 1.8].
• Each of the angles $GEF$ and $HEF$ (are) thus right angles [Def. 1.10] .
• Thus, $FE$ is at right angles to $GH$, which was drawn at random through $E$ (in the reference plane passing though $AB$ and $AC$).
• So, similarly, we can show that $FE$ will make right angles with all straight lines joined to it which are in the reference plane.
• And a straight line is at right angles to a plane when it makes right angles with all straight lines joined to it which are in the plane [Def. 11.3] .
• Thus, $FE$ is at "right angles to the reference plane":bookofproofs$2212. • And the reference plane is that (passing) through the straight lines$AB$and$CD$. • Thus,$FE$is at right angles to the plane (passing) through$AB$and$CD\$.
• Thus, if a straight line is set up at right angles to two straight lines cutting one another, at the common point of section, then it will also be at right angles to the plane (passing) through them (both).
• (Which is) the very thing it was required to show.

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