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Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another,

[Construction. EF is parallel to BC.

[I. 27. Wherefore the straight line EAF is drawn through the giden point A, parallel to the given straight line BC. Q.E.P.

PROPOSITION 32. 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 together equal to two right angles.

Let ABC be a triangle, and let one of its sides BC be produced to D: the exterior angle ACD shall be equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, namely, ABC,

Through the point Cdraw CE parallel to AB. [I. 31.

Then, because AB is parallel to CE, and AC falls on them, the alternate angles B

Ć BAC, ACE are equal.

[I. 29. Again, because AB is parallel to CE, and BD falls on them, the exterior angle ECD is equal to the interior and opposite angle ABC.

[I. 29. 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, ABC. [Axiom 2.

To each of these equals add the angle ACB; therefore the angles ACD, ACB are equal to the three angles CBA, BAC, ACB.

[Axiom 2. But the angles ACD, ACB are together equal to two right angles ;

[1. 13, therefore also the angles CBA, BAC, ACB are together equal to two right angles.

[Axiom 1. Wherefore, if a side of any triangle &c. Q.E.D.

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COROLLARY 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 sider.

For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles.

And by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as the figure has sides. And the same angles are equal to the interior angles of the figure, together with the angles at the point F, which

A is the common vertex of the triangles, that is, together with four right angles. [I. 15. Corollary 2.. Therefore all the interior angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

COROLLARY 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 angle ABD, is equal to two right angles;

[1. 13. therefore all the interior angles of the figure, together with all its exterior angles, are equal to D twice as many right angles as the figure has sides. But, by the foregoing Corollary all the interior angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. Therefore all the interior angles of the figure, together with all its exterior angles, 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,

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PROPOSITION 33. THEOREM. The straight lines which join the extremities of two

are also themselves equal and parallel.

Let AB and CD be equal and parallel straight lines, and let them be joined towards the same parts by the straight lines AC and BD: AC and BD shall be equal and parallel.

Join BC.

Then because AB is parallel to CD, . (Hypothesis. and BC meets them, the alternate angles ABC, BCD are equal. [I. 29. And because AB is equal to CD,

[Hypothesis. and BC is common to the two triangles ABC, DCB; the two sides AB, BC are equal to the two sides DC, CB, each to each; and the angle ABC was shewn to be equal to the angle BCD; therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles, each to each, to which the equal sides are opposite;

[I. 4. therefore the angle ACB is equal to the angle CBD.

And because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD.

[I. 27. And it was shewn to be equal to it. Wherefore, the straight lines &c. Q.E.D.

PROPOSITION 34. THEOREM. The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects the parallelogram, that is, divides it into two equal parts.

Note. A parallelogram is a four-sided figure of which the opposite sides are parallel; and a diameter is the straight line joining two of its opposite angleg.

Let ACDB be a parallelogram, of which BC is a diameter ; the opposite sides and angles of the figure sball be equal to one another, and the diameter BC shall bisect it.

Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to one another.

[I. 29. And because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal to one another.

[I. 29. Therefore the two triangles ABC, BCD have two angles ABC, BCA in the one, equal to two angles DCB, CBD in the other, each to each, and one side BC is 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, namely, the side AB equal to the side CD, and the side AC equal to the side BD, and the angle BAC equal to the angle CDB.

[I. 26. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD. JAx. 2. And the angle BAC has been shewn to be equal to the angle CDB. Therefore the opposite sides and angles of a parallelogram are equal to one another.

Also the diameter bisects the parallelogram. For AB being equal to CD, and BC common, the two sides AB, BC are equal to the two sides DC, CB each to each ; and the angle ABC has been shewn to be equal to the angle BCD; therefore the triangle ABCis equal to the triangle BCD,[1.4. and the diameter BC divides the parallelogram ACDB into two equal parts.

Wherefore, the opposite sides &c. Q.E.D.

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PROPOSITION 35. THEOREM. Parallelograms on the same base, and between the same parallels, are equal to one another.

Let the parallelograms ABCD, EBCF be on the same base BC, and between the same parallels AF, BC: the parallelogram ABCD shall be equal to the parallelogram EBCF. If the sides AD, DF of

A__ D the parallelograms ABCD, DBCF, opposite to the base BC, be terminated at the same point D, it is plain that each of the parallelograms is double of the triangle BDC;

[I. 34. and they are therefore equal to one another. [Axiom 6.

But if the sides AD, EF, opposite to the base BC of the parallelo

FA ED grams ABCD, EBCF be not terminated at the same point, then, because ABCD is a parallelogram AD is equal to BC;

[I. 34. for the same reason EF is equal to BC; therefore AD is equal to EF;

[Axiom 1. therefore the whole, or the remainder, AE is equal to the whole, or the remainder, DF.

[Axioms 2, 3. And AB is equal to DC;

[I. 34. therefore the two sides EA, AB are equal to the two sides FD, DC each to each; and the exterior angle FDC is equal to the interior and opposite angle EAB;

[I. 29. therefore the triangle EAB is equal to the triangle FDC.

[I. 4. Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB, and the remainders are equal; that is, the parallelogram ABCD is equal to the parallelogram EBCF.

Wherefore, parallelograms on the same base &c. Q.E.D.

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