If there are two equal plane angles, and on their vertices there are set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points are taken at random and perpendiculars are drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines are joined to the vertices of the original angles, then they contain with the elevated straight lines equal angles.

Let the angles *BAC* and *EDF* be two equal rectilinear angles, and from the points *A* and *D* let the elevated straight lines *AG* and *DM* be set up containing, with the original straight lines, equal angles respectively, namely, the angle *MDE* equal to the angle *GAB* and the angle *MDF* equal to the angle *GAC.*

Take the points *G* and *M* at random on *AG* and *DM.* Draw *GL* and *MN* from the points *G* and *M* perpendicular to the plane through *BA* and *AC* and the plane through *ED* and *DF,* and let them meet the planes at *L* and *N.* Join *LA* and *ND.*

I say that the angle *GAL* equals the angle *MDN.*

Make *AH* equal to *DM,* and draw *HK* through the point *H* parallel to *GL.*

Since *GL* is perpendicular to the plane through *BA* and *AC,* therefore *HK* is also perpendicular to the plane through *BA* and *AC.*

Draw *KC, NF, KB* and *NE* from the points *K* and *N* perpendicular to the straight lines *AC, DF, AB,* and *DE.* Join *HC, CB, MF,* and *FE.*

Since the square on *HA* equals the sum of the squares on *HK* and *KA,* and the sum of the squares on *KC* and *CA* equals the square on *KA,* therefore the square on *HA* equals the sum of the squares on *HK, KC,* and *CA.*

But the square on *HC* equals the sum of the squares on *HK* and *KC,* therefore the square on *HA* equals the sum of the squares on *HC* and *CA.* Therefore the angle *HCA* is right. For the same reason the angle *DFM* is also right.

Therefore the angle *ACH* equals the angle *DFM.* But the angle *HAC* equals the angle *MDF.* Therefore *MDF* and *HAC* are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that opposite one of the equal angles, that is, *HA* equals *MD,* therefore they also have the remaining sides equal to the remaining sides
respectively. Therefore *AC* equals *DF.*

Similarly we can prove that *AB* also equals *DE.* Since then *AC* equals *DF,* and *AB* equals *DE,* the two sides *CA* and *AB* equal the two sides *FD* and *DE.* But the angle *CAB* also equals the angle *FDE,* therefore the base *BC* equals the base *EF,* the triangle equals the triangle, and the remaining angles to the remaining angles. Therefore the angle *ACB* equals the angle *DPE.*

But the right angle *ACK* equals the right angle *DFN,* therefore the remaining angle *BCK* equals the remaining angle *EFN.* For the same reason the angle *CBK* also equals the angle *FEN.*

Therefore *BCK* and *EFN* are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that adjacent to the equal angles, that is, *BC* equals *EF,* therefore the remaining sides equal the remaining sides. Therefore *CK* equals *FN.*

But *AC* also equals *DF,* therefore the two sides *AC* and *CK* equal the two sides *DF* and *FN,* and they contain right angles. Therefore the base *AK* equals the base *DN.*

And, since *AH* equals *DM,* therefore the square on *AH* equals the square on *DM.* But the sum of the squares on *AK* and *KH* equals the square on *AH,* for the angle *AKH* is right, and the sum of the squares on *DN* and *NM* equals the square on *DM,* for the angle *DNM* is right, therefore the sum of the squares on *AK* and *KH* equals the sum of the squares on *DN* and *NM.* And of these the square on *AK* equals the square on *DN,* therefore the remaining square on *KH* equals the square on *NM.* Therefore *HK* equals *MN.*

And, since the two sides *HA* and *AK* equal the two sides *MD* and *DN* respectively, and the base *HK* equals the base *MN,* therefore the angle *HAK* equals the angle *MDN.*

Therefore, *if there are two equal plane angles, and on their vertices there are set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points are taken at random and perpendiculars are drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines are joined to the vertices of the original angles, then they contain with the elevated straight lines equal angles.*

Q.E.D.

From this it is clear that, *if there are two equal plane angles, and if elevated straight lines set up on them which are equal and contain equal angles with the original straight lines respectively, then the perpendiculars drawn from their ends to the planes in which are the original angles equal one another.*

Euclid’s proof is quite long. Various authors have substituted shorter ones.