# Proof: By Euclid

• Let $BAC$ and $EDF$ be two equal rectilinear angles.
• And let the raised straight lines $AG$ and $DM$ have been stood on points $A$ and $D$, containing equal angles respectively with the original straight lines. (That is) $MDE$ (equal) to $GAB$, and $MDF$ (to) $GAC$.
• And let the random points $G$ and $M$ have been taken on $AG$ and $DM$ (respectively).
• And let the $GL$ and $MN$ have been drawn from points $G$ and $M$ perpendicular to the planes through $BAC$ and $EDF$ (respectively).
• And let them have joined the planes at points $L$ and $N$ (respectively).
• And let $LA$ and $ND$ have been joined.
• I say that angle $GAL$ is equal to angle $MDN$. • Let $AH$ be made equal to $DM$.
• And let $HK$ have been drawn through point $H$ parallel to $GL$.
• And $GL$ is perpendicular to the plane through $BAC$.
• Thus, $HK$ is also perpendicular to the plane through $BAC$ [Prop. 11.8].
• And let $KC$, $NF$, $KB$, and $NE$ have been drawn from points $K$ and $N$ perpendicular to the straight lines $AC$, $DF$, $AB$, and $DE$.
• And let $HC$, $CB$, $MF$, and $FE$ have been joined.
• Since the (square) on $HA$ is equal to the (sum of the squares) on $HK$ and $KA$ [Prop. 1.47], and the (sum of the squares) on $KC$ and $CA$ is equal to the (square) on $KA$ [Prop. 1.47], thus the (square) on $HA$ is equal to the (sum of the squares) on $HK$, $KC$, and $CA$.
• And the (square) on $HC$ is equal to the (sum of the squares) on $HK$ and $KC$ [Prop. 1.47].
• Thus, the (square) on $HA$ is equal to the (sum of the squares) on $HC$ and $CA$.
• Thus, angle $HCA$ is a right angle [Prop. 1.48].
• So, for the same (reasons), angle $DFM$ is also a right angle.
• Thus, angle $ACH$ is equal to (angle) $DFM$.
• And $HAC$ is also equal to $MDF$.
• So, $MDF$ and $HAC$ are two triangles having two angles equal to two angles, respectively, and one side equal to one side - (namely), that subtending one of the equal angles - (that is), $HA$ (equal) to $MD$.
• Thus, they will also have the remaining sides equal to the remaining sides, respectively [Prop. 1.26].
• Thus, $AC$ is equal to $DF$.
• So, similarly, we can show that $AB$ is also equal to $DE$.
• Therefore, since $AC$ is equal to $DF$, and $AB$ to $DE$, so the two (straight lines) $CA$ and $AB$ are equal to the two (straight lines) $FD$ and $DE$ (respectively).
• But, angle $CAB$ is also equal to angle $FDE$.
• Thus, base $BC$ is equal to base $EF$, and triangle ($ACB$) to triangle ($DFE$), and the remaining angles to the remaining angles (respectively) [Prop. 1.4].
• Thus, angle $ACB$ (is) equal to $DFE$.
• And the right angle $ACK$ is also equal to the right angle $DFN$.
• Thus, the remainder $BCK$ is equal to the remainder $EFN$.
• So, for the same (reasons), $CBK$ is also equal to $FEN$.
• So, $BCK$ and $EFN$ are two triangles having two angles equal to two angles, respectively, and one side equal to one side - (namely), that by the equal angles - (that is), $BC$ (equal) to $EF$.
• Thus, they will also have the remaining sides equal to the remaining sides (respectively) [Prop. 1.26].
• Thus, $CK$ is equal to $FN$.
• And $AC$ (is) also equal to $DF$.
• So, the two (straight lines) $AC$ and $CK$ are equal to the two (straight lines) $DF$ and $FN$ (respectively).
• And they enclose right angles.
• Thus, base $AK$ is equal to base $DN$ [Prop. 1.4].
• And since $AH$ is equal to $DM$, the (square) on $AH$ is also equal to the (square) on $DM$.
• But, the (sum of the squares) on $AK$ and $KH$ is equal to the (square) on $AH$.
• For angle $AKH$ (is) a right angle [Prop. 1.47].
• And the (sum of the squares) on $DN$ and $NM$ (is) equal to the square on $DM$.
• For angle $DNM$ (is) a right angle [Prop. 1.47].
• Thus, the (sum of the squares) on $AK$ and $KH$ is equal to the (sum of the squares) on $DN$ and $NM$, of which the (square) on $AK$ is equal to the (square) on $DN$.
• Thus, the remaining (square) on $KH$ is equal to the (square) on $NM$.
• Thus, $HK$ (is) equal to $MN$.
• And since the two (straight lines) $HA$ and $AK$ are equal to the two (straight lines) $MD$ and $DN$, respectively, and base $HK$ was shown (to be) equal to base $MN$, angle $HAK$ is thus equal to angle $MDN$ [Prop. 1.8].
• Thus, if there are two equal rectilinear angles, and so on of the proposition.
• (Which is) the very thing it was required to show.

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