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Cor. 1.-If two lines KL and CD make, with EF, the two

angles KGH and GHC, together less than two right angles, KG and CH will meet on the side of EF on which the two

angles are that are less than two right angles. (Dem.) For, if not, KL and CD are either parallel, or they meet on the other side of EF; but they are not parallel ; for the angles KGH and GHC would then be together equal to two right angles. Neither do they meet toward the points L and D; for the angles LGH and GHD would then be two. angles of a triangle, and therefore less than two right angles; but this is impossible; for the four angles KGH, HGL, CHG, and GHD are together equal to four right angles, of which the two KGH and CHG are by supposition less than two right angles; therefore the other two, HGL and GHD, are greater than two right angles. Therefore, since KL and CD are not parallel, and do not meet towards L and D, ., they will meet if produced towards K and C. Cor. 2.-If two straight lines be each perpendicular to the

same line, they are parallel. For the interior angles on the same side are together equal to two right angles.

** * PROPOSITION Xxx. THEOREM. Straight lines which are parallel to the same straight line are parallel to one another.

Given that AB and CD are each of them parallel to EF; to prove that AB is also parallel to CD.

(Const.) Let the straight line GHK cut AB, EF, CD; and (Dem.) because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal to the angle GHF (I. 29).

B Again, because the straight line GK cuts F_ SH the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD; .c

and it was shewn that the angle AGK is equal to the angle GHF; therefore also AGK is equal to GKD (Ax. 1), and they are alternate angles; therefore AB is parallel to CD (I. 27).

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PROPOSITION XXXI. PROBLEM. To draw a straight line through a given point parallel to a given straight line. ,

Given the point A and the straight line BC; it is required to draw a straight line through the point À parallel to the straight line BC.

(Const.) In BC take any point D, and join AD; and at

the point A; in the straight line AD make (I. 23) the angle DAE
equal to the angle ADC; and produce
the straight line EA to F.
(Dein.) Because the straight line AD,

line AD' EMA f
which meets the two straight lines BC and
EF, makes the alternate angles EAD, -

BC ADC equal to one another, EF is parallel to BC (I. 27). Therefore the straight line EAF is drawn through the given point A parallel to the given straight line BC.

PROPOSITION XXXII. THEOREM. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles.

Given a triangle ABC, having one of its sides BC produced to Dto prove that the exterior angle ACD is equal to the two interior and opposite angles CAB and ABC, and that the three interior angles of the triangle-, namely, ABC, BCA, and CAB, are together equal to two right angles.

(Const.) Through the point C draw CE parallel to the straight line AB (I. 31). &

(Dem.) And because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (I. 29). Again, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC; but the angle ACE was shewn to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB and ABC. To each of these equals add the angle ACB, and the two angles ACD and ACB are equal to the three angles CBA, BAC, and ACB; but the angles ACD and ACB are equal to two right angles (I. 13); therefore also the angles CBA, BAC, and ACB are equal to two right angles. Cor. 1.--All the interior angles of any rectilineal figure,

together with four right angles, are equal to twice as many

right angles as the figure has sides. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has, sides, by drawing straight lines from a point within the figure to each of its pf angles. And, by the preceding proposition, l all the angles of these triangles are equal to : twice as many right angles as there are : triangles—that is, as there are sides of the figure; and the same angles are equal to the angles of the

figure, together with the angles at the point 0, which is the common vertex of the triangles; that is (I. 15, Cor. 2), together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Let the sum of the interior angles be denoted by I, the number of sides by n, and a right angle by R, then

I + 4R = 2nR. Cor. 2.--All the exterior angles of any rectilineal figure are

together equal to four right angles. Because every interior angle ABC, with its adjacent exterior ABD, is equal to two right angles (I. 13); therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles;

therefore all the exterior angles are equal to four right angles. Cor. 3.-If the sum of two angles in one triangle is equal to

the sum of two angles in another triangle, their third angles

must be equal. Cor. 4.-The two oblique angles of a right-angled triangle are together equal to a right angle.

PROPOSITION XXXIII. THEOREM. The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel.

Given AB and CD, two equal and parallel straight lines, joined towards the same parts by the straight lines, AC and BD; to prove that .AC and BD are also equal and parallel.

(Const.) Join BC; (Dem.) and because AB is parallel to CD, and BC meets them, the alternate angles ABC and BCD are equal (I. 29); and because AB is equal to CD, and BC common to the two triangles ABC and DCB, the two sides AB and BC are equal to the two DC and CB; and the angle ABC is equal to the angle DCB; therefore the base AC is equal to the base BD (I. 4), and the triangle ABC to the triangle DCB, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore the angle ACB is equal to the angle DBC; and because the straight line BC meets the two straight lines AC and BD, and makes the alternate angles ACB and DBC equal to one another, AC is parallel to BD (I. 27); and it was shewn to be equal to it. COR.-If two equal perpendiculars be drawn to a straight

line on the same side of it, the straight line joining their extremities is parallel to the former, and equal to the

intercepted part of it. For the perpendiculars are equal and also parallel (I. 28); therefore the lines joining their extremities are equal and parallel.

PROPOSITION XXXIV. THEOREM. The opposite sides and angles of a parallelogram are equal to one another, and the diagonal bisects it; that is, divides it in two equal parts.

Given a parallelogram ACDB, of which BC is a diagonal; to prove that the opposite sides and angles of the figure are equal to one another, and that the diagonal BC bisects it. (Dem.) Because AB is parallel to CD, and BC meets them,

the alternate angles ABC and DCB are equal to one another (I. 29); and because AC is parallel to BD, and BC meets them, the alternate angles ACB and DBC are equal to one another; wherefore the two triangles ABC and DCB have two angles ABC and BCA, in one, respectively equal to two angles DCB and CBD, in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other (I. 26)

namely, the side AB equal to the side CD, and AC to BD, and the angle BAC equal to the angle BDC. And because the angle ABC is equal to the angle DCB, and the angle CBD to the angle BCA, the whole angle ABD is equal to the whole angle ACD. And the angle BAC has been shewn to be equal to the angle BDC; therefore the opposite sides and angles of a parallelogram are equal to one another; also, the diagonal bisects it; for it was proved that the two triangles ABC and BCD are every way equal (I. 26), consequently the diagonal BC divides the parallelogram ACDB into two equal parts.

Cor. 1.-Parallel lines are equidistant.
For, if from two points in one of them perpendiculars be drawn

to the other, they are parallel (I. 28), and the two lines intercepted between them are parallel; therefore a parallelogram is formed, of which the perpendiculars are opposite sides, and are therefore equal. COR. 2.-Hence two triangles or two parallelograms between

the same parallels have the same altitude; and if they have the same altitude, and be on the same side of bases that are in the same straight line, they are between the same parallels. (I. 33, Cor.)

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PROPOSITION XXXV. TAEOREM. Parallelograms upon the same base, and between the same parallels, are equal to one another.

Given two parallelograms ABCD and EBCF (see the 2d and' 3d figures) upon the same base BC, and between the same parallels AF, BC; to prove that the parallelogram ABCD is equal to the parallelogram EBCF.

(Dem.) If the sides ÅD and DF, of the parallelograms ABCD, DBCF, oppo

co" I site to the basę BC, be terminated in the same point D (as in fig. 1), it is plain that each of the : parallelograms is double of the triangle DBC (I. 34);. 'and** they are therefore equal to one another.

But if the sides AD and EF, opposite to the base BC of the parallelograms ABCD and EBCF, be not terminated in the same point, then because ABCD is a parallelogram, AD is equal to 1 BC; for the same reason EF is equal to BC; wherefore, AD is equal to EF (Ax. 1); and DE is common; there. I fore (Ax. 2 or 3) the whole or the remainder AE is equal to the whole or the remainder DF; AB also is equal to DC; and the two EA and AB are therefore equal to the two A D £ FAELE FD and DC, each to each; and the exterior angle FDC is equal (I. 29) to the interior EAB; therefore the base EB is B equal to the base FC, and the triangle. EAB equal to the triangle FDC (I. 4); take the triangle FDC from the trapezoid ABCF, and from the same trapezoid take the triangle EAB, the remainders therefore are equal; that is (Ax. 3), the parallelogram ABCD is equal to the parallelogram EBCF.

PROPOSITION XXXVI. THEOREM. Parallelograms upon equal bases, and between the same "parallels, are equal to one another.

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