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

PROP. XXVI. THEOREM.

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

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

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

Then, because AB is parallel to EF (by Hyp.), and GH interfects them, the angle AGH is equal to the alternate angle GHF (Prop. 24.)

And becaufe CD is parallel to EF (by Hyp.), and HK interfecs them, the outward angle GHF is equal to the inward angle HKD (Prop. 25.)

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

And, fince the right line GK interfects the two right lines AB, CD, and makes the angle AGK equal to the alternate angle GKD, AB will be parallel to CD, as was to be fhewn.

PRO P. XXVII. PROBLEM.

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

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Let AB be the given right line, and c the given point; it is required to draw a right line through the point c that shall 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.

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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.

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PRO P. XXVIII. THEOREM.

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.

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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 EE is parallel to AC, and CB interfects them, the angle CBE will be equal to the alternate angle BCA (Prop. 24.)

And because BE is parallel to AC, and AD 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 these 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.); consequently the three angles BCA, CAB and ABC, taken together, are alfo equal to two right angles.

COROLL. I. 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.

PRO P. XXIX. THEOREM.

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

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Let AB, DC be two equal and parallel right lines; then will the right lines AD, BC, which join the corresponding extremes of those lines, be alfo equal and parallel.

For draw the diagonal, or right line AC:

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

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And, because AB is equal to DC (by Hyp.), AC common to each of the triangles ABC, ADC, and the angle DCA equal to the angle CAB, the fide AD will also be equal to the fide BC, and the angle DAC to the angle ACB (Prop. 4.)

Since, therefore, the right line AC interfects the two right lines AD, BC, and makes the alternate angles equal to each other, thofe lines will be parallel (Prop.23.)

But the line AD has been proved to be equal to the line BC; consequently they are both equal and parallel. QE.D.

PROP. XXX. THEOREM.

The oppofite fides and angles of any parallelogram are equal to each other, and the diagonal divides it into two equal parts.

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Let ABCD be a parallelogram, whose diagonal is AC; then will its oppofite fides and angles be equal to each other, and the diagonal AC will divide it into two equal parts.

For, fince the fide AD is parallel to the fide BC (Def. 22.), and the right line AC interfects them, the angle DAC will be equal to the alternate angle ACB (Prop. 24.)

And, because the fide DC is parallel to the fide AB (Def. 22.), and AC interfects them, the angle DCA will be equal to the alternate angle CAB (Prop. 24.)

Since, therefore, the two angles DAC, DCA, are equal to the two angles ACB, CAB, each to each, the remain

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