PROPOSITION XVI.

If, in a plane, a straight line, meeting two other straight lines, make the alternate angles equal to one another, these two straight lines shall be parallel.

Let the straight line GH, meeting the two straight lines PQ, RS, make the alternate s HGP, GHS equal to one another, and therefore also the alternate LS HGQ, GHR equal to one another. (I. 14, Cor.) Then shall PQ, RS be parallel.

For if not, PQ, RS when produced will meet either towards and S or towards R and P.

Let them, if possible, meet when produced towards Q and S in the point X.

Thus GHX is a A, and if it were applied to GH on the opposite side, then HX would fall along GP, because HGP is equal to GHS; and GX would fall along HR, because GHR is equal to

HGQ.

Therefore GP, HR would meet if produced towards P and R, and thus two straight lines would enclose a space, which is impossible.

Therefore PQ, RS will not meet if produced towards Q and S.

Similarly it may be shown that they will not meet if produced towards P and R.

Therefore PQ and RS are parallel.

Wherefore if in a plane a straight line &c.

Q.E.D.

PROPOSITION XVII.

To draw a straight line through a given point parallel to a given straight line.

A

B

C

Let Fbe the given point, and CD the given straight line.

It is required to draw a straight line through F parallel to CD.

Join Fwith any point G in CD.

At the point Fin FG make ▲ GFA equal to ▲ FGD, and produce AF to B.

(I. 10)

Then shall AB be parallel to CD.

For since FG, meeting the two straight lines AB and CD, makes the alternate angles GFA, FGD equal to one another,

therefore AB is parallel to CD.

Wherefore a straight line has been drawn &c.

(I. 16)

Q.E.F.

PROPOSITION XVIII.

If a straight line meet two parallel straight lines, the alternate angles shall be equal to one another.

A
I

Let the straight line FG meet the two parallel straight lines AB and CD.

Then shall GFA be equal to the alternate FGD.

For, if possible, let them be unequal, and at Fin FG make GFX equal to ▲ FGD.

Then XF is parallel to CD.

(I. 10)

(I. 16)

Therefore through F two straight lines have been

drawn parallel to CD, which is impossible.

Therefore GFA is not unequal to FGD-that is,

it is equal to it.

Wherefore if a straight line &c.

Q.E.D.

COROLLARY 1.-If a straight line fall upon two parallel straight lines, the exterior angle shall be equal to the interior and opposite angle upon the same side of the line.

Let HK fall upon the two parallel straight lines AB,

CD;

then shall HFB be = L FGD.

For AFG is the alternate L FGD;

=

==

and AFG is the vertical opposite & HFB; (I. 11) therefore

HFB is: = L FGD.

COROLLARY 2.-If a straight line fall upon two parallel straight lines, the two interior angles upon the same side of the line shall together be equal to two right angles.

That is, the s BFG, FGD shall be together equal to two right angles.

For AFG is the alternate L FGD;

=

=

therefore the LS BFG, AFG are together the s BFG, FGD;

but the LS BFG, AFG are together = two right angles; (I. 14) therefore S BFG, FGD are together two right

=

angles.

PROPOSITION XIX.

If one side of a triangle be produced, the exterior angle shall be equal to the two interior and opposite angles.

and let one of its sides BC be produced to K.

Then shall the exterior angle ACK be equal to the two interior and opposite angles BAC, ABC.

Through C draw CH parallel to AB.

(I. 17)

Now because CH is parallel to AB, and AC meets them,

=

.. the ACH is the alternate L BAC. (I. 18) Again, because CH is parallel to AB, and BK falls upon them;

.. the exterior HCK is the interior and opposite ABC; (I. 18, Cor.) . the whole ACK is the two angles BAC, ABC together.

Wherefore if one side of a triangle &c.

Q.E.D.