SPECIFIC SUBJECTS.

MATHEMATICS.-SECOND STAGE.

EUCLID.-Book I,

XV.-If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

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

Because the straight line AE makes with CD the angles CEA, AED, these angles are together equal to two right angles. Again, because

the straight line DE makes with AB
the angles AED, DEB, these also are A
together equal to two right angles;
and CEA, AED, have been shown to

be equal to two right angles; wherefore the angles CEA, AED, are equal to AED, DEB. Take away the common angle AED, and the remaining angle DEA is equal to DEB. In the same manner it can be demonstrated that the angles CEB, AEB, are equal.

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

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

XVI.-If one side of a triangle be produced, the exterior

angle is greater than either of the interior opposite angles.

Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles CBA, BAC.

Bisect AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G.

E

Because AE is equal to EC, and BE to EF; AE, EB, are equal to CE, EF, each to each; and the angle AEB is equal to CEF, because they are opposite vertical angles; therefore the base AB is equal to CF, and the triangle AEB to CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal

to ECF; but the angle ECD is greater than ECF; therefore the angle ACD is greater than BAE; in the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG, that is the angle ACD, is greater than ABC.

XVII.—Any two angles of a triangle are together less than two right angles.

Let ABC be any triangle; any two of its angles are together less than two right angles. Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater than the interior and opposite angle ABC; to each of these add the angle ACB; therefore the angles ACD, ACB, are greater than ABC, ACB; but

B

C

ACD, ACB, are together equal to two right angles, therefore ABC, BCA, are less than two right angles. In like manner it may be demonstrated that BAC, ACB,as also CAB, ABC, are less than two right angles. XVIII.—The greater side of every triangle is opposite to the greater angle.

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 BCA.

D

Because AC is greater than AB, make AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, it is greater than the interior and opposite angle DCB; but ADB is equal to ABD, because the side AB is equal to AD; therefore the angle ABD is likewise greater than ACB. Wherefore much more is the angle ABC greater than ACB.

XIX,―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 BCA; the side AC is likewise greater than the side AB.

For if it be not greater, AC must either be equal to AB or less than it : it is not equal, because then the angle ABC would be equal to ACB; but it is not; therefore AC is not equal to

B

AB; neither is it less; because then the angle ABC would be less than ACB; but it is not; therefore the side AC is not less than AB; and has been shown that it is not cqual to AB; therefore it is greater than AB.

XX.-Any two sides of a triangle are together greater than the third side.

Let ABC be a triangle; any two sides of it together

are greater than the third side, viz., the sides BA, AC, greater than the side BC; and AB, BC, greater than AC, and BC, CA, greater than AB. Produce BA to point D, and make AD equal to AC; and join DC. Because DA is equal to AC, the angle ADC is likewise equal to ACD; but the angle BCD is greater than ACD; therefore the angle BCD is greater than ADC; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater side is opposite to the

greater angle; therefore the side DB is greater than BC; but DB is equal to BA and AC; therefore BA, AC, are greater than BC. In the same manner it may be shown that AB, BC, are greater than CA; and BC, CA, greater than AB.

XXI.-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 the two straight lines BD, CD, be drawn from BC, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides BA, AC, of the triangle, but contain an angle BDC greater than the angle BAC.

Produce BD to E; and because two B

sides of a triangle are greater than the third side, the two sides BA, AE, of the triangle ABE are greater than BE. To each of these add EC; therefore BA, AC, are greater than BE, EC; again, because the two sides CE, ED, of the triangle CED are greater than CD, add DB to each of these; therefore CE, EB, are greater than CD, DB; but it has been shown that BA, AC, are greater than BE, EC, much more then are BA, AC, greater than BD, DC.

Again, because the exterior angle of a triangle is

greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been shewn that the angle BDC is greater than CEB; much more then is BDC greater than BAC. XXII. To make a triangle of which the sides shall be

K

equal to three given straight lines, but any two whatever of these must be greater than the third. Let A, B, C, be the three given straight lines, of which any two whatever are greater than the third viz., A and B greater than C; A and C greater than B; and B and C than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each to each.

D

F G HE

B.

C

Take a straight line DE terminated at the point, D, but unlimited towards E, and make DF equal to A, FG to B, and GH equal to C; and from the centre F, at the distance FD, describe the circle DKL; and from the centre G, at the distance GH, describe another circle HLK; and join KF, KG; the triangle KFG has its sides equal to the three straight lines, A,B,C.

Because the point F is the centre of the circle DKL, FD is equal to FK; but FD is equal to the straight line A; therefore FK is equal to A. Again, because G is the centre of the circle LKH, GH is equal to GK; but GH is equal to C; therefore also GK is equal to C; and FG is equal to B; therefore the three straight lines KF, FG, GK, are equal to the three A, B, C; and therefore the triangle KFG has its three sides KF, FG, GK, equal to the three given straight lines A, B, C. XXIII.—At a given point in a given straight line, to make α rectilineal angle equal to a given rectilineal

angle.

Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle; it is