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Let the two straight lines AB, CD cut one another at the point E; the angle AEC shall be equal to the angle DEB, and the angle CEB to the angle AED.

Because the straight line AEmakes with the straight line CD the angles CEA, AED, these angles are together equal to two right angles.

[I. 13. Again, because the straight line DE makes with the straight line AB the angles AED, DEB, these also are together equal to two right angles.

. [I. 13. But the angles CEA, AED have been shewn to be together equal to two right angles. Therefore the angles CEÀ, AED are equal to the angles AED, DEB. From each of these equals take away the common angle AED, and the remaining angle CEA is equal to the remaining angle DEB.

Axiom 3. In the same manner it may be shewn that the angle CEB is equal to the angle AED.

Wherefore, if two straight lines &c. Q.E.D.

Corollary 1. From this it is manifest that, if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles.

Corollary 2. And consequently, that all the angles made by any number of straight lines meeting at one point, are together equal to four right angles.

PROPOSITION 16. THEOREM. If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles.

Let ABC be a triangle, and let one side BC be produced to D: the exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC. Bisect AC at E,

[I. 10. join BE and produce it to F, making EF equal to EB, [I. 3. and join FC.

Because AE is equal to EC, and BE to EF; [Constr. the two sides AE, EB are equal to the two sides CE, EF, each to each ;

and the angle AEB is equal to the angle CEF,
because they are opposite ver-
tical angles ;

[I. 15. therefore the triangle AEB is equal to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite;

[I. 4. therefore the angle BAE is equal to the angle ECF. But the angle ECD is greater than the angle ECF. [Axiom 9. Therefore the angle ACD is greater than the angle BAE.

In the same manner if BC be bisected, and the side AC be produced to G, it may be shewn that the angle BCG, that is the angle ACD, is greater than the angle ABC. [I. 15. Wherefore, if one side &c. Q.E.D.

PROPOSITION 17. THEOREM. Any two angles of a triangle are togсther less than two right angles.

Let ABC be a triangle: any two of its angles are together less than two right angles.

Produce BC to D.

Then because ACD is the exterior angle of the triangle ABC, it is greater than the interior opposite angle ABC.

[I. 16. To each of these add the angle ACB Therefore the angles ACD, ACB are greater than the angles ABC, ACB. But the angles ACD, ACB are together equal to two right angles.

[I. 13. Therefore the angles ABC, ACB are together less than two right angles.

In the same manner it may be shewn that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles.

Wherefore, any tro angles &c. Q.E.D.

PROPOSITION 18. THEOREM. The greater side of every triangle has the greater angle opposite to it.

Let ABC be a triangle, of which the side AC is greater than the side AB: the angle ABC is also greater than the angle ACB.

Because AC is greater than AB, make AD equal to AB, [I. 3. and join BD.

Then, because ADB is the exterior angle of the triangle BDC, it is greater than the interior op- $ posite angle DCB. [I. 16. But the angle ADB is equal to the angle ABD, [I. 5. because the side AD is equal to the side AB. [Constr. Therefore the angle ABD is also greater than the angle ACB. Much more then is the angle ABO greater than the angle ACB.

[Axiom 9. Wherefore, the greater side &c. Q.E.D.

PROPOSITION 19. THEOREM. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it,

Let ABC be a triangle, of which the angle ABC is greater than the angle ACB: the side AC is also greater

For if not, AC must be either equal to AB or less than AB. But AC is not equal to AB, for then the angle ABC would be equal to the angle ACB; [1.5. but it is not; (Hypothesis. therefore AC is not equal to AB. Neither is AC less than AB, for then the angle ABC would be less than the angle ACB;

[I. 18. but it is not ;

[Hypothesis.

therefore AC is not less than AB.
And it has been shewn that AC is not equal to AB.
Therefore AC is greater than AB.

Wherefore, the greater angle &c. Q.E.D.

PROPOSITION 20. THEOREM. Any two sides of a triangle are together greater than the third side.

Let ABC be a triangle: any two sides of it are together greater than the third side; namely, BA, AC greater than BC; and AB, BC greater than AC; and BC, CA greater than AB.

Produce BA to D, making AD equal to AC,

[I. 3. and join DC.

Then, because AD is equal to AC, [Construction. the angle ADC is equal to the angle ACD.

[I. 5. But the angle BCD is greater than the angle ACD. [AX. 9. Therefore the angle BCD is greater than the angle BDC. And because the angle BCD of the triangle BCD is greater than its angle BDC, and that the greater angle is subtended by the greater side;

[I. 19. therefore the side BD is greater than the side BC. But BD is equal to BA and AC. Therefore BA, AC are greater than BC.

In the same manner it may be shewn that AB, BC are greater than AC, and BC, CA greater than AB.

Wherefore, any two sides &c. Q.E.D.

PROPOSITION 21. THEOREM. If from the ends of the side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.

Let ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to the point D within the triangle: BD, DC shall be less than the other two sides BA, AC of the triangle, but shall contain an angle BDC greater than the angle BAC.

Produce BD to meet AC at E.

Because two sides of a triangle are greater than the third side, the two sides BA, AE of the triangle ABE are greater than the side BE.

[I. 20. To each of these add EC. Therefore BA, AC are greater than BE, EC. Again ; the two sides CE, ED of the triangle CED are greater than the third side CD.

[I. 20. To each of these add DB. Therefore CE, EB are greater than CD, DB. But it has been shewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC.

Again, because the exterior angle of any triangle is greater than the interior opposite angle, the exterior angle BDC of the triangle CDE is greater than the angle CED.

[I. 16. For the same reason, the exterior anglo CEB of the triangle ABE is greater than the angle BAE. But it has been shewn that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC.

Wherefore, if from the ends &c. Q.E.D.

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