base GH is equal to the base KL, and the angle HGв equal to the angle BKL:

And the angles GBM, KBN are equal, since they are opposite vertical angles; therefore the two triangles GBM, KBN have the two angles MGB, MBG respectively equal to the two angles BKN, NBK, and the sides BG, BK adjacent to the equal angles in each equal. Therefore these two triangles are equal in every respect (i. 26); and hence the side GM is equal to the side KN, and the side MB to the side NB:

Again BG is equal to BK, BA common to the two triangles ABG, ABK, and the angles ABG, ABK are equal (Def. 10), because AB by hyps is at right angles to CD; therefore these two triangles have the two sides GB, BA respectively equal to the two sides KB, BA and the included angle GBA equal to the included angle KBA. fore they are equal in every respect; and hence the base AG is equal to the base AK:

There

And in like manner from the pair of equal triangles ABH, ABE it may be shewn that the base AH is equal to the base AL:

Also GH was proved to be equal to KL; therefore the two triangles AGH, AKL have the three sides AG, GH, HA respectively equal to the three sides AK, KL, LA. Therefore these two triangles are equal in every respect (i. 8); and hence the angle AGH is equal to the angle AKL:

And AG is equal to AK, and GM to KN; therefore the two triangles AGM, AKN have the two sides AG, GM respectively equal to the two sides AK, KN, and the included angle AGM equal to the included angle AKN. Therefore these

two triangles are equal in every respect; and hence the base AM is equal to the base AN:

And lastly, MB was shewn to be equal to BN, and AB is common to the two triangles ABM, ABN: therefore these two triangles have the three sides, AM, MB, BA respectively equal to the three sides AN, NB, BA. Therefore they are equal in every respect; and hence the angle ABM is equal to the angle ABN:

Thus AB standing on MN makes with it the adjacent angles ABM, ABN equal to one another; therefore by the def of a right angle each of these angles is a right angle, and AB is at right angles to MBN. In like manner it may be shewn that AB is at right angles to every other straight line drawn meeting it in the plane passing through CD, EF: GL, HK being joined instead of GH, KL, when the straight line lies between BG and BE, and BF and BD, instead of between BC and BF, and EB and BD as in the figure: therefore by the def" of a straight line being perpendicular to a plane (xi. Def. 3), AB is perpendicular to the plane passing through CD, EF. Which was to be proved.

PROP. V. THEOR.

If a straight line be at right angles to each of three straight lines at the point where they meet: then these three straight lines shall be in one plane.

Let AB be at right angles to each of the three straight lines BC, BD, BE at the point в where they meet. Then BC, BD, BE shall be in one plane.

For if they are not all in one plane; then, since each two must be in one plane because they cut one another (xi. 2), any two will be in one plane, and the third without it. Let, if possible, BD, BE be in one plane, while BC is without this plane through BD, BE; and through AB, BC draw a plane cutting the plane through BD, BE in the common section BF, which is a straight line (xi. 3), and

which cannot coincide with BC, because BC is supposed without the plane through BD, BE.

By the const" BA, BC, BF are all in one plane, viz. that through BA, BC. Now because AB is at right angles to each of the straight lines BD, BE

C

F

B

E

at the point of their intersection B, it is perpendicular to the plane through them (xi. 4); and BF is drawn meeting it in that plane: therefore by the defn of a straight line being perpendicular to a plane (xi. Def. 3), AB is at right angles to BF, and ABF a right angle. But ABC is a right angle by hyps; and all right angles are equal (Ax. 10): therefore the angle ABF is equal to the angle ABC. Now, they are both in one plane, that through BA, BC, so that ABC is a part of the whole ABF; hence the whole ABF is equal to the part ABC: which is impossible (Ax. 9). Therefore no one of the three straight lines BC, BD, BE can be without the plane in which the other two are; that is, the three BC, BD, BE are all in one plane. Which was to be proved.

PROP. VI. THEOR.

If two straight lines be each of them perpendicular to the same plane: then they shall be parallel to one another.

Let the two straight lines, AB, CD be perpendicular to the same plane FG at the points B, D. Then AB shall be parallel to CD.

Join BD, which by the def" of a plane (Def. 7) is in the plane FG; in this plane draw (i. 11) DH at right angles to BD, and from DH cut off (i. 3) DE equal to AB.

BE, AE, AD.

Join

Because AB is perpendicular to the plane FG by hyps; and DB, EB are drawn meeting it in that plane: therefore by the def" of a straight line being perpendicular to a

plane (xi. Def. 3), each of the angles ABD, ABE is a right angle. For like reason each of the angles CDB, CDE is a right angle. Now because AB is equal to DE by const", BD common to the two triangles ABD, BDE, and the right angle

ABD equal to the right angle BDE, since all right angles are equal (Ax. 11); therefore these two triangles have the two sides AB, BD respectively equal to the two sides ED, DB and the included angle ABD equal to the included angle EDB. Therefore they are equal in every respect (i. 4) ; and hence the base AD is equal to the base BE. And AB is equal to DE, and AE common to the two triangles ABE, ADE; therefore these two triangles have the three sides AB, BE, EA respectively equal to the three sides ED, DA, AE. Therefore they are equal in every respect (i. 8); and hence the angle ABE is equal to the angle EDA. But ABE

is a right angle; therefore also EDA is a right angle. And EDB, EDC are both right angles; hence ED is at right angles to each of the three straight lines DB, DA, DC at the point D where they meet. Therefore these three straight lines are all in one plane (xi. 5); and AB is in the plane in which BD, DA are, because each of the three straight lines AB, BD, DA cuts the other two (xi. 2): therefore AB, BD, DC are all in one plane. Now each of the angles ABD, BDC is a right angle; and hence BD cutting the two straight lines AB, CD, which have been proved to be in the same plane with it, in B and D makes the two interior angles ABD, CDB on the same side of BD together equal to two right angles: therefore AB is parallel (i. 28) to CD. Which was to be proved.

If two straight lines be parallel: then the straight line drawn joining any point in the one with any point in the other shall be in the same plane with the two parallels.

Let AB, CD be two parallel straight lines; E any point in AB; F any point in CD. Then the straight line EF, which joins E and F, shall be in the same plane with AB, CD.

For if not let it, if possible, be without the plane and have the position EGF, and in the plane in which the parallels AB, CD are draw (Post. 1) the straight line EHF from E to F, which does not coincide with EGF.

are each straight lines; that is, two straight

F

D

lines enclose a space: which is impossible (Ax. 10). Hence EF cannot be without the plane in which AB, CD are, that is, it is in it. Which was to be proved.

OBS. This prop" may be proved at once from the defn of a plane; for since E, F are points in the plane in which the parallels AB, CD are, therefore EF which joins them lies in this plane (Def. 7).

PROP. VIII. THEOR.

If two straight lines be parallel, and one of them be perpendicular to a plane: then the other also shall be perpendicular to that plane.

Let AB, CD be two parallel straight lines, meeting the plane FG in the points B, D; and let one of them AB be perpendicular to the plane. Then the other CD shall also be perpendicular to the plane FG.

Join BD, which by the def" of a plane (Def. 7) is in the plane FG; in this plane draw (i. 11) DH at right angles to BD; and from DH cut off (i. 3) DE equal to AB. Join BE, AE, ad.