Sidebilder
PDF
ePub

PROPOSITION XX.

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

[blocks in formation]

Let the straight line GH, meeting the two straight lines PQ, RS, make the alternates PGH, GHS equal to one another.

Then shall PQ be || to RS.

For, if not, PQ, RS when produced will meet either towards QS or towards PR.

Let them, if possible, be produced to meet towards Q, S in the point X.

Then GHX is a ;

and the exterior ▲ PGH is > the interior and opposite L GHX;

but it is also equal to it, which is impossible.

(I. 13)

.. PQ, RS do not meet when produced towards Q, S. Similarly it may be proved that they will not meet if produced towards P, R.

.. they are parallel.

Hence are easily deduced the following Theorems :

If a straight line, falling upon two other straight lines. makes the exterior angle equal to the interior and opposite upon the same side of the line, the two straight lines shall be parallel.

[blocks in formation]

Let the straight line EFGH, falling upon the two straight lines AB, CD, make the exterior angle EFB = the interior

[blocks in formation]
[blocks in formation]

▲ FGD, and these are alternate angles; .. AB is to CD.

Also

If a straight line falling upon two other straight lines makes the interior angles upon the same side of the line together equal to two right Ls, these two straight lines shall be parallel.

For let the 4s BFG, FGD be together equal to right ▲ s; .. they are together =LS BFG, AFG:

[blocks in formation]

(1.9)

L FGD is = L AFG, and these are alternate angles;

.. AB is to CD.

PROPOSITION XXI.

If a straight line meets two parallel straight lines it makes the alternate angles equal to one another.

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

For if not, if possible, at F in FG draw another line XFY, making L XFG = L FGD.

Then XFY is || to CD.

(I. 20)

.. through F two straight lines have been drawn || to CD, which is impossible..

.. LAFG is the alternate FGD.
=

(Lemma.)

Hence are easily deduced the following Theorems :

If a straight line fall upon two parallel straight lines, it makes the exterior L= the interior and opposite upon the same side.

[blocks in formation]

And AFG is the vertically opposite & HFB; (1.6)

=

.. LHFB is = L FGD.

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

That is, the S BFG, FGD shall be together = two right angles.

For AFG is = the FGD,

and.. S AFG, BFG are togethers FGD, BFG. But s AFG, BFG are together two rights; (1.9) two rights.

:. Ls FGD, BFG are together

=

=

COR. If AB, BC are two straight lines respectively || to DE, EF, then shall ▲ ABC be = L DEF.

[blocks in formation]

For, let BC, ED, produced, if necessary, intersect in H.

Then Ls ABC, DEF are each = ▲ DHG, and are therefore equal to one another.

PROPOSITION XXII.

Straight lines which are parallel to the same straight line are parallel to one another.

A

B

Let the straight lines A, B be each of them || to C.
Then shall A, B be || to one another.

For if not they will meet if produced, and there will thus be drawn through one point two straight lines parallel to the same straight line, which is impossible. (Lemma.)

« ForrigeFortsett »