---
book: 11
number: 35
id: "XI.35"
kind: "theorem"
uses: ["[[book-11/proposition-8]]", "[[book-1/proposition-47]]", "[[book-1/proposition-48]]", "[[book-1/proposition-26]]", "[[book-1/proposition-4]]", "[[book-1/proposition-8]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.35"
license: "CC-BY-SA-4.0"
---

# XI.35

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

## Proof

Let the angles *BAC*, *EDF* be two equal rectilineal angles, and from the points *A*, *D* let the elevated straight lines *AG*, *DM* be set up containing, with the original straight lines, equal angles respectively, namely, the angle *MDE* to the angle *GAB* and the angle *MDF* to the angle *GAC*, let points *G*, *M* be taken at random on *AG*, *DM*, let *GL*, *MN* be drawn from the points *G*, *M* perpendicular to the planes through *BA*, *AC* and *ED*, *DF*, and let them meet the planes at *L*, *N*, and let *LA*, *ND* be joined; I say that the angle *GAL* is equal to the angle *MDN*.

Let *AH* be made equal to *DM*, and let *HK* be drawn through the point *H* parallel to *GL*.

But *GL* is perpendicular to the plane through *BA*, *AC*; therefore *HK* is also perpendicular to the plane through. *BA*, *AC*. [[book-11/proposition-8|XI. 8]]

From the points *K*, *N* let *KC*, *NF*, *KB*, *NE* be drawn perpendicular to the straight lines *AC*, *DF*, *AB*, *DE*, and let *HC*, *CB*, *MF*, *FE* be joined.

Since the square on *HA* is equal to the squares on *HK*, *KA*, and the squares on *KC*, *CA* are equal to the square on *KA*, [[book-1/proposition-47|I. 47]] therefore the square on *HA* is also equal to the squares on *HK*, *KC*, *CA*.

But the square on *HC* is equal to the squares on *HK*, *KC*; [[book-1/proposition-47|I. 47]] therefore the square on *HA* is equal to the squares on *HC*, *CA*.

Therefore the angle *HCA* is right. [[book-1/proposition-48|I. 48]]

For the same reason the angle *DFM* is also right.

Therefore the angle *ACH* is equal to the angle *DFM*.

But the angle *HAC* is also equal to the angle *MDF*.

Therefore *MDF*, *HAC* are two triangles which have 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*; therefore they will also have the remaining sides equal to the remaining sides respectively. [[book-1/proposition-26|I. 26]]

Therefore *AC* is equal to *DF*.

Similarly we can prove that *AB* is also equal to *DE*.

Since then *AC* is equal to *DF*, and *AB* to *DE*, the two sides *CA*, *AB* are equal to the two sides *FD*, *DE*.

But the angle *CAB* is also equal to the angle *FDE*; therefore the base *BC* is equal to the base *EF*, the triangle to the triangle, and the remaining angles to the remaining angles; [[book-1/proposition-4|I. 4]] therefore the angle *ACB* is equal to the angle *DFE*.

But the right angle *ACK* is also equal to the right angle *DFN*; therefore the remaining angle *BCK* is also equal to the remaining angle *EFN*.

For the same reason the angle *CBK* is also equal to the angle *FEN*.

Therefore *BCK*, *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* equal to *EF*; therefore they will also have the remaining sides equal to the remaining sides. [[book-1/proposition-26|I. 26]]

Therefore *CK* is equal to *FN*.

But *AC* is also equal to *DF*; therefore the two sides *AC*, *CK* are equal to the two sides *DF*, *FN*; and they contain right angles.

Therefore the base *AK* is equal to the base *DN*. [[book-1/proposition-4|I. 4]]

And, since *AH* is equal to *DM*, the square on *AH* is also equal to the square on *DM*.

But the squares on *AK*, *KH* are equal to the square on *AH*, for the angle *AKH* is right; [[book-1/proposition-47|I. 47]] and the squares on *DN*, *NM* are equal to the square on *DM*, for the angle *DNM* is right; [[book-1/proposition-47|I. 47]] therefore the squares on *AK*, *KH* are equal to the squares on *DN*, *NM*; and of these the square on *AK* is equal to the square on *DN*; therefore the remaining square on *KH* is equal to the square on *NM*; therefore *HK* is equal to *MN*.

And, since the two sides *HA*, *AK* are equal to the two sides *MD*, *DN* respectively, and the base *HK* was proved equal to the base *MN*, therefore the angle *HAK* is equal to the angle *MDN*. [[book-1/proposition-8|I. 8]]

Therefore etc.

Porism. From this it is manifest that, if there be two equal plane angles, and if there be set up on them elevated straight lines which are equal and contain equal angles with the original straight lines respectively, the perpendiculars drawn from their extremities to the planes in which are the original angles are equal to one another. Q. E. D.