PROP. XXVIÍ.

PROBLÉM,

Through a given point, to draw a right line parallel to a given right line.

Let AB be the given right line, and c the given point; it is required to draw à right line through the point c that fhall be parallel to AB:

Take any point D in AB, and make DE equal to DC (Prop. 3.); and from the points C, E, with the distances CD, ED, defcribe the arcs rs, nm.

Then, fince any two fides of the triangle ECD are, together, greater than the third fide (Prop. 18.), thofe arcs will interfect each other (Prop. 19.)

Let them interfect at F; and through the points F, C, draw the line GH, and it will be parallel to AB, as was required.

For, fince the fides CF, FE of the triangle EFC are each equal to the fide CD, or DE, of the triangle CDE, (by Conft.) and EC is common, the angle ECF will be equal to the angle CED (Prop. 7.)

But these are alternate angles; therefore GH is parallel to AB (Prop. 24.); and it is drawn through the point c, as was to be done.

COROLL. If a right line intersect two other right lines, and make the two inward angles, on the fame fide, together lefs than two right angles, thofe lines, being produced, will meet each other.

Right lines which are parallel to the fame right line, are parallel to each other.

A

E

Let the right lines AB, CD be each of them paralle EF, then will AB be parallel to CD.

For, draw any right line GK, cutting the lines AB CD, in the points G, H and K.

Then, because AB is parallel to EF (by Hyp.), an intersects them, the angle AGH is equal to the alt angle GHF (Prop. 24.)

And because CD is parallel to EF (by Hyp.), a interfe&ts them, the outward angle GHF is equa inward angle HKD (Prop. 25.)

But the angle AGH has been fhewn to be equ angle GHF; therefore the angle AGK is also equ angle GKD.

And, fince the right line K interfects the lines AB, CD, and makes the angle AGK eq alternate angle GKD, AB will be parallel to c to be fhewn.

PROP. XXVIII. THEOREM.

A

If one fide of a triangle be produced, the outward angle will be equal to the two inward oppofite angles, taken together; and the three angles of every triangle, taken together, are equal to two right angles.

Let ABC be a triangle, having one of its fides AB produced to D; then will the outward angle CBD be equal to the two inward oppofite angles BCA, CAB, taken together; and the three angles BCA, CAB and ABC, taken together, are equal to two right angles.

For through the point B, draw the right line BE parallel to AC (Prop. 28.)

Then, because BE is parallel to AC, and CB întersects them, the angle CBE will be équal to the alternate angle BCA (Prop. 24.)

And because BE is parallel to AC, and Ap interfects them, the outward angle EBD will be equal to the inward angle CAB (Prop. 25.)

But the angles CBE, EBD are equal to the whole angle CBD; therefore the outward angle CBD is equal to the two inward oppofite angles BCA, CAB taken together.

And if, to thefe equals, there be added the angle ABC, the angles CBD, ABC, taken together, will be equal to the three angles BCA, CAB and ABC, taken together.

But the angles CBD, ABC, taken, together, are equal to two right angles (Prop. 13.); confequently the three angles BCA, CAB and ABC, taken together, are alfo equal to two right angles.

COROLL. 1. If two angles of one triangle, be equal to two angles of another, each to each, the remaining angles will also be equal.

COROLL, 2. Any quadrilateral may be divided into two triangles; therefore all the four angles of fuch a figure, taken together, are equal to four right angles.

PROP.

XXIX.

THEOREM.

Right lines joining the correfponding extremes of two equal and parallel right lines are themselves equal and parallel.

Let AB, DC be two equal and parallel right lines; ther will the right lines AD, BC, which join the correfponding extremes of thofe lines, be alfo equal and parallel.

For draw the diagonal, or right line AC :

Then, because AB is parallel to DC, and AC interfects them, the angle DCA will be equal to the alternate angle CAB (Prop. 24.)