hcju/proofwikiqa huggingface dataset:
:300px Suppose it were possible to have a straight line in more than one plane. Let a part $AB$ of the straight line $ABC$ be in the plane of reference, and another part $BC$ be in a plane more elevated. There will then be in the plane of reference some straight line $BD$ continuous with $AB$ in a straight line. Therefore $AB$ is a common segment of the two straight lines $ABC$ and $ABD$. Suppose a circle is described with center $B$ and radius $AB$. Then the diameters would cut off unequal arcs of the circle. Hence the result. {{qed}} {{Euclid Note|1|XI}}
:400px Let the two straight lines $AB$ and $CD$ intersect at the point $E$. Let $F$ and $G$ be arbitrary points on $EC$ and $EB$. Let $CB$ and $FG$ be connected. Let $FH$ and $GK$ be drawn across. Suppose part of $\triangle ECB$ is in the plane of reference and the rest of it in another plane. Then a part of one of the straight lines $EC$ and $EB$ will be in the plane of reference and the result in another plane. But from {{EuclidPropLink|book = XI|prop = 1|title = Straight Line cannot be in Two Planes}}, this cannot happen. Therefore $\triangle ECB$ is all in one plane. But whatever plane $\triangle ECB$ is in, each of the straight lines $EC$ and $EB$ are also in that plane. And in whatever plane $EC$ and $EB$ are in, the straight lines $AB$ and $CD$ are also in that plane. Therefore $AB$ and $CD$ are in one plane, and every triangle is in one  plane. {{qed}} {{Euclid Note|2|XI}}
:300px Let $AB$ and $BC$ be two distinct planes that cut one another. Let $DB$ be their common section. Suppose $DB$ were not a straight line. Then let: : the straight line segment $DEB$ be drawn in the planes $AB$ and: : the straight line segment $DFB$ be drawn in the planes $BC$. Thus the two straight line segments $DEB$ and $DFB$ have the same endpoints. Thus $DEB$ and $DFB$ enclose an area, which is absurd. Therefore $DEB$ and $DFB$ are not straight lines. Hence the result. {{qed}} {{Euclid Note|3|XI}}
:400px 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 {{EuclidPropLink|book = I|prop = 15|title = 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 {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$\angle DAE = \angle EBC$ But from {{EuclidPropLink|book = I|prop = 15|title = 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 {{EuclidPropLink|book = I|prop = 26|title = Triangle Angle-Side-Angle and Side-Angle-Angle Equality}}: :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 {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$FA = FB$ 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 {{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}}: :$\angle FAD = \angle FBC$ Similarly: :$AG = BH$ and: :$FA = FB$ and so $FA = AG$ and $FB = BH$. We also have that: :$\angle FAG = \angle FBH$ and so from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$FG = FH$ 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 {{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}}: :$\angle GEF = \angle HEF$ 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 {{EuclidDefLink|XI|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. {{qed}} {{Euclid Note|4|XI}}
:400px Let $AB$ be a straight line set up at right angles to the three straight lines $BC$, $BD$ and $BE$ at their point of meeting $B$. It is to be demonstrated that $BC$, $BD$ and $BE$ are all in the same plane. Suppose they are not all in the same plane, but that while $BD$ and $BE$ are in the plane of reference, $BC$ is in a plane more elevated. Thus the plane through $AB$ and $BC$ is different from the plane of reference. From {{EuclidPropLink|book = XI|prop = 3|title = Common Section of Two Planes is Straight Line}}: :the common section of the plane through $AB$ and $BC$ with the plane of reference is a straight line $BF$, say. Therefore the three straight lines $AB$, $BC$ and $BF$ are in the same plane, that is, the one through $AB$ and $BC$. We have that $AB$ is at right angles to each of the straight lines $BD$ and $BE$. Therefore from {{EuclidPropLink|book = XI|prop = 4|title = Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane}}: :$AB$ is at right angles to the plane through $BD$ and $BE$. But the plane through $BD$ and $BE$ is the plane of reference. Therefore $AB$ is at right angles to the plane of reference. Thus from {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}: :$AB$ will also make right angles with all the straight lines which meet it and are in the plane of reference. But $BF$, which is in the plane of reference, meets $AB$. Therefore $\angle ABF$ is a right angle. But by hypothesis $\angle ABC$ is also a right angle. Therefore $\angle ABF = \angle ABC$. But $\angle ABF$ and $\angle ABC$ are in one plane, which is impossible. Therefore $BC$ is not in a plane more elevated. Therefore $BC$, $BD$ and $BE$ are all in the same plane. {{qed}} {{Euclid Note|5|XI}}
:400px Let $AB$ and $CD$ be two straight lines at right angles to the plane of reference. It is to be demonstrated that $AB$ is parallel to $CD$. Let $AB$ and $CD$ meet the plane of reference at $B$ and $D$ respectively. Let the straight line $BD$ be joined. Let $DE$ be drawn in the plane of reference at right angles to $BD$ such that $DE = AB$. Let $BE$, $AE$ and $AD$ be joined. We have that $AB$ is at right angles to the plane of reference. So by {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}: :$AB$ is at right angles to every straight line which meets it and is in the plane of reference. But each of $BD$ and $BE$ is in the plane of reference and meets $AB$. Therefore $\angle ABD$ and $\angle ABE$ are both right angles. For the same reason $\angle CDB$ and $\angle CDE$ are both right angles. We have that $AB = DE$, and that $BD$ is common. Thus the two sides $AB$ and $BD$ of $\triangle ABD$ equal the two sides $BD$ and $BE$ of $\triangle EDB$. The triangles $\triangle ABD$ and $\triangle EDB$ both include right angles. Therefore by {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$AD = BE$ We have that: :$AB = DE$ and: :$AD = BE$ Thus the two sides $AB$ and $BE$ of $\triangle ABE$ equal the two sides $ED$ and $DA$ of $\triangle EDA$. We also have that $AE$ is common. So from {{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}}: :$\angle ABE = \angle EDA$ But $\angle ABE$ is a right angle. Therefore $ED$ is at right angles to $DA$. But $ED$ is also at right angles to the straight lines $BD$ and $DC$. Therefore $ED$ is set up at right angles to the three straight lines $BD$, $DA$ and $DC$ at their meeting points. Therefore from {{EuclidPropLink|book = XI|prop = 5|title = Three Intersecting Lines Perpendicular to Another Line are in One Plane}}: :$BD$, $DA$ and $DC$ are in the same plane. But from {{EuclidPropLink|book = XI|prop = 2|title = Two Intersecting Straight Lines are in One Plane}}: :the triangle $DAB$ is in one plane. Therefore $AB$ is in the same plane as $DB$ and $DA$. Therefore the straight lines $AB$, $BD$ and $DC$ are in one plane. Also, each of $\angle ABD$ and $\angle BDC$ is a right angle. Therefore from Supplementary Interior Angles implies Parallel Lines: :$AB$ is parallel to $CD$. {{qed}} {{Euclid Note|6|XI}}
:350px Let $AB$ and $CD$ be two straight lines which are parallel. Consider the plane in which they lie to be the plane of reference. Let $E$ and $F$ be taken at random on both of them. Then the straight line $EF$ is to be demonstrated to lie in the same plane as $AB$ and $CD$. Suppose to the contrary that $EF$ is in a more elevated plane: $EGF$. Let a plane be drawn through $EGF$. From {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}, the common section of this plane with the plane of reference is a straight line passing through $E$ and $F$. Therefore $EF$ and $EGF$ enclose an area, which is impossible. Therefore the straight line $EF$ joining $E$ and $F$ is in the same plane as $AB$ and $CD$. {{qed}} {{Euclid Note|7|XI}}
:400px Let $AB$ and $CD$ be two parallel straight lines. Let $AB$ be at right angles to the plane of reference. Then $CD$ needs to be demonstrated to be also at right angles to the plane of reference. Let $AB$ and $CD$ meet the plane of reference at $B$ and $D$. Let $BD$ be joined. Then by {{EuclidPropLink|book = XI|prop = 7|title = Line joining Points on Parallel Lines is in Same Plane}}: :$AB$, $CD$ and $BD$ are all in the same plane. Let $DE$ be drawn in the plane of reference at right angles to $BD$ and equal to $AB$. Let $BE$, $AE$ and $AD$ be joined. We have that $AB$ is at right angles to the plane of reference. Therefore from {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}, $AB$ is at right angles to all the straight lines which meet it and are in the plane of reference. Therefore each of $\angle ABD$ and $\angle ABE$ is a right angle. We have that the straight line $BD$ falls on the parallels $AB$ and $CD$. Therefore from Parallelism implies Supplementary Interior Angles $\angle ABD$ and $\angle CDB$ are equal to two right angles. But $\angle ABD$ is a right angle. Therefore $\angle CDB$ is also a right angle. Therefore $CD$ is at right angles to $BD$. We have that $AB = DE$, and that $BD$ is common. The two sides $AB$ and $BD$ of $\triangle ABD$ are equal to the two sides $ED$ and $DB$ of $\triangle EBD$. Also $\angle ABD = \angle EDB$, as they are both right angles. Therefore $AD = BE$. We have that: :$AB = DE$ and: :$BE = AD$ Thus the two sides $AB$ and $BE$ of $\triangle ABE$ are equal to the two sides $ED$ and $DA$ of $\triangle EAD$. We have that $AE$ is common. So: : $\angle ABE = \angle EDA$ But $\angle ABE$ is a right angle. Therefore $\angle EDA$ is a right angle. Therefore $ED$ is at right angles to $AD$. But $ED$ is also at right angles to $DB$. Therefore by {{EuclidPropLink|book = XI|prop = 4|title = Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane}}: :$ED$ is at right angles to the plane through $DA$ and $BD$. But from {{EuclidPropLink|book = XI|prop = 2|title = Two Intersecting Straight Lines are in One Plane}}: :$DC$ is in the plane through $DA$ and $BD$ inasmuch as $AB$ and $BD$ are in the same plane which $AB$ and $BD$ are. Also $DC$ is in the same plane which $AB$ and $BD$ are. Therefore $ED$ is at right angles to $DC$. So $CD$ is also at right angles to $DE$. But $CD$ is also at right angles to $BD$. Therefore $CD$ is set up at right angles to the straight lines $DE$ and $DB$ which cut one another, at $D$. So from {{EuclidPropLink|book = XI|prop = 4|title = Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane}}: :$CD$ is also at right angles to the plane through $DE$ and $DB$ are. But the plane through $DE$ and $DB$ is the plane of reference. Hence the result. {{qed}} {{Euclid Note|8|XI}}
:450px Let $AB$ and $CD$ be straight lines which are both parallel to another straight line $EF$ which is in a different plane. It is to be demonstrated that $AB$ is parallel to $CD$. Let $G$ be an arbitrary point on $EF$. Let $GH$ be drawn at right angles to $EF$, in the plane holding both $AB$ and $EF$. Let $GK$ be drawn at right angles to $EF$, in the plane holding both $CD$ and $EF$. We have that $EF$ is at right angles to each of the straight lines $GH$ and $GK$. Therefore from {{EuclidPropLink|book = XI|prop = 4|title = Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane}}: :$EF$ is at right angles to the plane through $GH$ and $GK$. We have that $EF$ is parallel to $AB$. Therefore from {{EuclidPropLink|book = XI|prop = 8|title = Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane}}: :$AB$ is at right angles to the plane through $GH$ and $GK$. For the same reason: :$CD$ is at right angles to the plane through $GH$ and $GK$. Therefore each of the straight lines $AB$ and $CD$ is at right angles to the plane through $GH$ and $GK$. From  {{EuclidPropLink|book = XI|prop = 6|title = Two Lines Perpendicular to Same Plane are Parallel}}: :$AB$ is parallel to $CD$. {{qed}} {{Euclid Note|9|XI}}
:450px Let $AB$ and $BC$ be straight lines which meet one another. Let $DE$ and $EF$ be straight lines which meet one another such that $AB$ is parallel to $DE$ and $BC$ is parallel to $EF$. It is to be demonstrated that $\angle ABC = \angle DEF$. Let $BA, BC, ED, EF$ be cut off equal to one another. Let $AC, CF, BE, AD, DF$ be joined. We have that: :$BA = ED$ and :$BA \parallel ED$ Therefore from {{EuclidPropLink|book = I|prop = 33|title = Lines Joining Equal and Parallel Straight Lines are Parallel}}: :$AD = BE$ and :$AD \parallel BE$ For the same reason: :$CF = BE$ and :$CF \parallel BE$ So each of $AD$ and $CF$ is equal and parallel to $BE$. But from {{EuclidPropLink|book = XI|prop = 9|title = Lines Parallel to Same Line not in Same Plane are Parallel to each other}}: :$AD = CF$ and :$AD \parallel CF$ We have that $AC$ and $DF$ join $AD$ and $CF$. Therefore from {{EuclidPropLink|book = I|prop = 33|title = Lines Joining Equal and Parallel Straight Lines are Parallel}}: :$AC = DF$ and :$AC \parallel DF$ So we have that $AB$ and $BC$ are equal and parallel to $DE$ and $EF$. We also have that $AC = DF$. So from {{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}}: :$\angle ABC = \angle DEF$ {{qed}} {{Euclid Note|10|XI}}
:500px Let $A$ be the given elevated point. Let the given plane be identified as the plane of reference. It is required that a straight line be drawn perpendicular to the plane of reference. Let any arbitrary straight line $BC$ be drawn in the plane of reference. From {{EuclidPropLink|book = I|prop = 11|title = Construction of Perpendicular Line}}: :Let $AD$ be drawn from $A$ perpendicular to $BC$. If $AD$ is perpendicular to the plane of reference, then the construction is complete. Otherwise, let $DE$ be drawn from $D$ perpendicular to $BC$ and in the plane of reference. From {{EuclidPropLink|book = I|prop = 12|title = Perpendicular through Given Point}}: :Let $AF$ be drawn from $A$ perpendicular to $DE$. From {{EuclidPropLink|book = I|prop = 31|title = Construction of Parallel Line}}: :Let $GH$ be drawn from the point $F$ parallel to $BC$. We have that $BC$ is perpendicular to each of $DA$ and $DE$. Therefore from {{EuclidPropLink|book = XI|prop = 4|title = Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane}}: :$BC$ is perpendicular to the plane through $ED$ and $DA$. We have that $GH$ is parallel to $BC$. But from {{EuclidPropLink|book = XI|prop = 8|title = Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane}}: :$GH$ is also perpendicular to the plane through $ED$ and $DA$. Therefore by {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}: :$GH$ is perpendicular to all the straight lines which meet it and are in the plane through $ED$ and $DA$. But $AF$ meets $GH$ and is in the plane through $ED$ and $DA$. Therefore $GH$ is perpendicular to $FA$. So $FA$ is perpendicular to $GH$. But $FA$ is also perpendicular to $DE$. Therefore $FA$ is perpendicular to both $GH$ and $DE$. Therefore from {{EuclidPropLink|book = XI|prop = 4|title = Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane}}: :$AF$ is perpendicular to the plane through $DE$ and $GH$. But the plane through $DE$ and $GH$ is the plane of reference. Therefore $AF$ is perpendicular to the plane of reference. Thus $AF$ is the required line. {{qed}} {{Euclid Note|11|XI}}
:400px Let the given plane be identified as the plane of reference. Let $A$ be the point from which the perpendicular is to be constructed. Let $B$ be any elevated point. From {{EuclidPropLink|book = XI|prop = 11|title = Construction of Straight Line Perpendicular to Plane from point not on Plane}}: :let $BC$ be constructed from $B$ perpendicular to the plane of reference. From {{EuclidPropLink|book = I|prop = 31|title = Construction of Parallel Line}}: :let $AD$ be drawn from $A$ parallel to $BC$. We have that $AD$ and $BC$ are two parallel straight lines of which one of them $BC$ is perpendicular to the plane of reference. From {{EuclidPropLink|book = XI|prop = 8|title = Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane}}: :$AD$ is also perpendicular to the plane of reference. Therefore $AD$ is the required perpendicular. {{qed}} {{Euclid Note|12|XI}}
:400px Suppose it were possible to set up two straight lines $AB$ and $AC$ perpendicular to the plane of reference and on the same side. Let a plane be drawn through $BA$ and $AC$. From {{EuclidPropLink|book = XI|prop = 3|title = Common Section of Two Planes is Straight Line}}: :let the common section be the straight line $DAE$. Therefore the straight lines $AB$, $AC$ and $DAE$ are all in the same plane. We have that $AC$ is perpendicular to the plane of reference. So from {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}: :$AC$ is perpendicular to all the straight lines which meet it and are in the plane of reference. But $DAE$ meets $AC$ and is in the plane of reference. Therefore $\angle CAE$ is a right angle. For the same reason, $\angle BAE$ is a right angle. Therefore $\angle BAE = \angle CAE$. But both are in the same plane, which is impossible. Hence the result. {{qed}} {{Euclid Note|13|XI}}
:500px Let $AB$ be a straight line which is perpendicular to each of the planes $CD$ and $EF$. It is to be demonstrated that $CD$ and $EF$ are parallel. Suppose, to the contrary, that $CD$ and $EF$ are not parallel. Then when produced they will meet. From {{EuclidPropLink|book = XI|prop = 3|title = Common Section of Two Planes is Straight Line}}: :let the common section be the straight line $GH$. Let $K$ be an arbitrary point on $GH$. Let $AK$ and $BK$ be joined. We have that $AB$ is perpendicular to $EF$. So from {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}: :$AB$ is perpendicular to the straight line $BK$ in the planes $EF$ produced. Therefore $\angle ABK$ is a right angle. For the same reason $\angle BAK$ is a right angle. Thus, in $\triangle ABK$, there are two angles which are right angles. From {{EuclidPropLink|book = I|prop = 17|title = Two Angles of Triangle Less than Two Right Angles}} this is impossible. Therefore $CD$ and $EF$ do not meet when produced. So from {{EuclidDefLink|XI|8|Parallel Planes}}: :$CD$ and $EF$ are parallel. {{qed}} {{Euclid Note|14|XI}}
:400px Let $AB$ and $BC$ be two straight lines which meet each other at $B$. Let $DE$ and $EF$ be two straight lines which meet each other at $E$ such that: :$AB$ is parallel to $DE$ :$BC$ is parallel to $EF$. It is to be demonstrated that the plane holding $AB$ and $BC$ is parallel to the plane holding $DE$ and $EF$. From {{EuclidPropLink|book = XI|prop = 11|title = Construction of Straight Line Perpendicular to Plane from point not on Plane}}: :let $BG$ be drawn from $B$ perpendicular to the plane holding $DE$ and $EF$. From {{EuclidPropLink|book = I|prop = 31|title = Construction of Parallel Line}}: :let $GH$ be drawn from $G$ parallel to $ED$ and :let $GK$ be drawn from $G$ parallel to $EF$. We have that $BG$ is perpendicular to the plane holding $DE$ and $EF$. Therefore from {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}:  :$BG$ is perpendicular to all the straight lines which meet it and are in the plane holding $DE$ and $EF$. But each of $GH$ and $GK$ meets $BG$ and is in the plane holding $DE$ and $EF$. Therefore each of $\angle BGH$ and $\angle BGK$ is a right angle. From {{EuclidPropLink|book = XI|prop = 9|title = Lines Parallel to Same Line not in Same Plane are Parallel to each other}}: :$BA$ is parallel to $GH$. Therefore from {{EuclidPropLink|book = I|prop = 29|title = Parallelism implies Supplementary Interior Angles}}: :$\angle GBA$ and $\angle BGH$ equal two right angles. But $\angle BGH$ is a right angle. Therefore $\angle GBA$ is a right angle. Therefore $GB$ is perpendicular to $BA$. For the same reason, $GB$ is also perpendicular to $BC$. We have that the straight line $GB$ is set up perpendicular to the two straight lines $BA$ and $BC$ which cut one another. Therefore from {{EuclidPropLink|book = XI|prop = 4|title = Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane}}: :$GB$ is perpendicular to the plane holding $AB$ and $BC$. But from {{EuclidPropLink|book = XI|prop = 14|title = Planes Perpendicular to same Straight Line are Parallel}}: :the plane holding $AB$ and $BC$ is parallel to the plane holding $DE$ and $EF$. {{qed}} {{Euclid Note|15|XI}}
:500px Let $AB$ and $CD$ be two planes which are parallel. Let $AB$ and $CD$ be cut by the plane $EFGH$. Let: :$EF$ be the common section of $EFGH$ and $AB$ and: :$GH$ be the common section of $EFGH$ and $CD$. It is to be shown that $EF$ is parallel to $GH$. Suppose that $EF$ and $GH$ are not parallel. Then when produced, they will meet, either in the direction of $F$ and $H$, or in the direction of $E$ and $G$. WLOG suppose they meet at $K$ when produced in the direction of $F$ and $H$. We have that $EFK$ is in the plane $AB$. Therefore from {{EuclidPropLink|book = XI|prop = 1|title = Straight Line cannot be in Two Planes}}: :all the points of $EFK$ are in the plane $AB$. But $K$ is one of the points of $EFK$. Therefore $K$ is in the plane $AB$. For the same reason, $K$ is also in the plane $CD$. Therefore the planes $AB$ and $CD$ will meet when produced. But they do not meet, because they are parallel. Therefore $EF$ and $GH$ will not meet when produced in the direction of $F$ and $H$. By the same argument, $EF$ and $GH$ will not meet when produced in the direction of $E$ and $G$. But from {{EuclidDefLink|I|23|Parallel Lines}}: :$EF$ is parallel to $GH$. {{qed}} {{Euclid Note|16|XI}}