And since the angles CBD, ABC are equal to two right angles (30), and 2 CBD is equal to the two angles BAC, ACB, .. the angles BAC, ACB, ABC, are equal to two right angles.

Another Method. That the three angles of a triangle are equal to two right angles may be shewn in another very simple way thus:

Let ABC be the triangle, and through C draw DCE parallel to AB (36), then the D angles ACB, ACD, BCE are together equal to two right angles (30. Cor. 2). But since is parallel to AB, LACD= 2 BAC, and <BCE= -ABC (34, Cor.4), :: the angles BAC, ACB, ABC, are equal to two right angles.

Cor. Hence no triangle can have more than one right angle, or one obtuse angle.

38. PROP. XVI. Any two sides of a triangle are together grealer than the third side.

[If this be not evident from the fact that the straight line joining any two points is less than any crooked line joining the same points, the following proof may be given:]

Let ABC be a triangle, produce AC to D, making CD equal to CB, (by drawing a circle with centre cand radius CB); and join BD.

Then, since CB=CD, CBD = = 2 CDB (26); but - ABD is greater than <CBD,.. Z ABD is greater than CDB, or ZADB; and in every Ā triangle the greater side is opposite to the greater angle, .: in the triangle ABD, AD is greater than AB; and AD= AC + CB, since CB=CD, .. AC + CB is greater than AB.

Similarly it may be shewn that AB + BC is greater than AC, and AC + AB greater than BC.

CoR. Hence, also, the difference between any two sides is less than the third side.

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39. PROP. XVII. If two triangles have two angles of the one equal to two angles of the other, each to each, and likewise the side which is common to those angles in the one equal to the side which is common to the two angles equal to them in the other, the triangles shall be equal in all respects.

Let ABC, DEF, be two triangles, in which 2 ABC = _DEF, ZACB = 2 DFE, and side A BC = side EF; then the triangle ABC shall be equal to the triangle DEF in all respects.

For, if the triangle ABC be “applied to,' or laid upon, the triangle DEF, so that the point B shall be upon E, and the line BC upon EF, the point C will fall upon F, be- B cause BC = EF. Also the side BA will fall upon ED, because - ABC= _DEF; and CA will fall upon FD, because . ACB =

= 2 DFE (8). Hence BA coinciding with ED, and CA with FD, the point A cannot but coincide with the point D; and .. the triangles coincide, or are equal in all respects.

E 40. Prop. XVIII. In every parallelogram the opposite sides are equal to one another; and so are the opposite angles. Likewise the diameter*, or diagonal, divides the parallelogram into two equal parts.

Let ACDB be a parallelogram, of which BC is a diameter, or diagonal; then AB=CD, AC = BD, 2 ABD = LACD, .BAC = <BDC, and the triangle ABC= the triangle BCD.

All that is here known and given is, that AB is parallel to CD, and AC parallel to BD (13).

Now since AB is parallel to CD, and BC meets them, LABC= _BCD (34, Cor. 4); and since AC is parallel

* The word diameter' is seldom used in this meaning, being almost wholly restricted to the . Circle'.


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to BD, and BC meets them, - ACB = - DBC; ... in the two triangles ABC, BCD, two angles of the one are equal to two angles of the other, each to each, and one side BC, viz. the side common to those angles, the same in both, the triangles are equal in all respects (39), and .. AB=CD, AC=BD, and the triangle ABC = the triangle BCD, that is, the diameter BC divides the parallelogram ACDB into two equal parts.

Also, since - ABC = 2 BCD, and 2 ACB= 2 DBC,

LABC + 2 DBC, or ABD = = BCD+ 2 ACB, or LACD. And since the triangles ABC, BCD, are equal in all respects, .. - BAC = - BDC.

Cor. 1. Hence, if any two adjacent sides of a parallelogram be equal to two adjacent sides of another parallelogram, each to each, and the angles contained by those sides are equal, the parallelograms will be equal in all respects.

COR. 2. The two diagonals of every parallelogram bisect each other.

Join AD, and let BC, AD, intersect in E; then, since - ABE = 2 DCE, and - BAE = - CDE, and side AB = side CD, the triangles ABE, CDE, are equal in all respects (39); •, side AE = side DE. Similarly it may be shewn that BE=CE; therefore AD, BC bisect each other in the point E.

41. Prop. XIX. All parallelograms, which have one side common and the sides opposite in one and the same straight line, are equal to one another*.

Let ABCD, EBCF, be any two parallelograms, having the side BC

E D common to both, and their opposite sides AD, EF in the same straight line AEDF; the parallelogram ABCD shall be equal to the parallelogram EBCF.

B * Euclid's enunciation of this is · Parallelograms upon the same base and between the same parallels are equal to one another'. By

base' is meant the side on which the parallelogram may be supposed to stand ; and between the same parallels' means that the parallelograms are bounded by those parallels in two directions.



For AD=BC = EF (40), taking DE from each of these equals, AE = DF ; also EB = FC (40), and 2 AEB = CFD (34, Cor. 4), since BE, CF are parallel ; .. in the triangles AEB, DFC, the two sides AE, EB are equal to the two sides DF, FC, each to each, and LAEB= < DFC, .. the triangles are equal in all respects (24), that is, the triangle AEB=the triangle DFC. Now, if these equal triangles be separately subtracted from the same area ABCF, the remainders must be equal, that is, the area ABCD= the area EBCF.

[Observe, when D falls between A and E, DE must be added, instead of subtracted.]

Cor. 1. Hence, also, all parallelograms upon equal bases and between the same two parallels' are equal to one another.

For, if ABCD, EFGH, be two parallelograms upon equal bases BC, FG,

I E K H and between the same parallels AH, BG, from the points F and G draw FI, GK, parallel to AB, or meeting AH, or AH B с produced, in I and K. Then since ABFI is a parallelogram, FI=AB (40). And FG=BC; and 2 GFI = -ABC, *. the parallelogram FGKI = the parallelogram ABCD, (40. Cor. 1). But the parallelogram EFGH = the parallelogram FGKI, as has been proved, .. the parallelogram EFGH=the parallelogram* ABCD.

Cor. 2. Join BD, and FH, then BD is a diagonal of the parallelogram ABCD, and divides it into two equal parts (40); .. the triangle BCD= half the parallelogram ABCD. Similarly the triangle FGH = half the parallelogram EFGH. And the halves of equal things must themselves be equal; .. the triangle BCD= the triangle FGH; that is, all triangles upon the same, or equal bases, and between the same parallels', are equal to one another.

Parallelogram' being a long word, when it occurs frequently, it is allowable to abridge it thus, om. The word parallel' also, in writing, is often abridged to ll. It is not advisable, however, for the learner to carry this sort of symbolism very far.

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CD (36),


42. PROP. XX. To describe or construct a square upon a given straight line, that is, so as to have the given straight line for one of its sides.

Let AB be the given straight line. From the point A draw the indefinite straight line AC at right angles to AB (28); with centre A and radius A B describe an arc of a circle cutting AC in D; through D draw DE parallel to AB, and through B draw BE parallel to AD: then ABED shall be a square.

For AB = AD; and ABED is a parallelogram, which has its opposite sides equal, and also its opposite angles (40); .. AB=DE = AD= BE, that is, all the sides of ABCD are equal. Also, since 2 BAD is a right angle, and AD is parallel to BE, .. ZABE is a right angle. But the opposite angles are equal, .. all the angles of ABED are right angles.

Hence, by the definition, ABED is a square, and it is described upon the line AB.

Cor. 1. If the side of one square be equal to the side of another square, the squares are equal in all respects.

Cor. 2. If a parallelogram have one of its angles a right angle, it has four right angles.

COR. 3. If two squares be equal, their sides are equal.

43. Prop. XXI. In the case of any right-angled triangle the square described upon the side opposite to the right angle* is equal to the two squares together described upon the two other sides which form the right angle.

Let ABC be a right-angled triangle, in which 2 ACB is the right angle. Upon the side AB, opposite to the right angle describe the square ADEB (42); upon BC the square BCFG; and upon AC the square AHKC. Then the square ADEB shall be equal to the two squares BCFG, and AHKC taken together (21).

The side of the triangle which is opposite to the right angle is sometimes called the hypothenuse', from a Greek word signifying to subtend because it subtends the right angle.

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