XXVIII.

If a line, AC, cross another, BD, in the point E, the opposite angles will be equal, viz BEA CED, and BEC=AED.

Upon the oint E as a centre, describe the circle ABCD; then it is evident that ABC is a semicircle, as also BCD (by Art. 6); therefore the arc ABC-arc BCD; taking from both the common arc BC, there remains arc AB=arc CD; that is, the angle BEA is equal to the angle CED. After the same manner we may prove that the angle BEC is equal to the angle AED.

XXIX.

B

If a line, GH, cross two parallel lines, AB, CD, it makes the external opposite angles equal to each other; viz. GEB⇒CFH, and AEG=HFD.

For since AB and CD are parallel to each other, they may be considered as one broad line, and GH crossing it; then the vertical or opposite angles, GEB, CFH, are equal (by Art. 28), as also AEG HFD.

XXX.

E B

A

C F

H

If a line, GH, cross two parallel lines, AB, CD (see the figure), the alternate angles, AEF and EFD, or CFE and FEB, are equal. For GEB= AEF (Art. 28), as also CFH EFD (by the same Art.), but GEB-CFH by the last; therefore AEF is equal to EFD; in the same way may we prove FEB⇒CFE.

XXXI.

If a line, GH, cross two parallel lines, AB, CD (see the preceding figure), the external angle, GEB, is equal to the internal opposite one, EFD, or AEG equal to CFE.

For the angle AEF is equal to the angle EFD by the last, and AEF = GEB (by Art. 28); therefore GEB EFD; in the same way we may prove AEG=CFE.

XXXII.

If a line, GH, cross two parallel lines, AB, CD (see the preceding figure), the sum of the two internal angles, BEF and DFE, or AEF and CFE, is equal to two right angles.

For since the angle GEB is equal to the angle EFD (by Art. 31), to both add the angle BEF, and we have GEB+BEF: BEFEFD; but GEB+ BEF = two right angles (Art. 27). Hence, BEF + EFD two right angles; and in the same manner we may prove AEF+CFE two right angles.

XXXIII.

In any triangle, ABC, one of its legs, as BC, being produced towards D, the externa. angle, ACD, is equal to the sum of the internal and opposite angles, ABC, BAC.

To prove this, through C draw CE parallel to AB; then, since CE is parallel to AB, and the lines AC, BD cross them, the angle ECD:= ABC (by Art. 31), and ACE=BAC (by Art. 30); adding these together we have ECD + ACE = ABC + BAC; but B ECD+ACE=ACD; therefore ACD = ABC + BAC.

XXXIV.

A

Hence it may be proved that if any two lines, AB and CD, be crossed by a third line, EF, and the alternate angles, AEF and EFD, be equal, the lines AB and CD will be varallel.

For, if they are not parallel, they must meet each other on one side of the line EF (suppose at G), and so form the triangle EGF, one of whose sides, GE, being produced to A, the exterior angle, AEF, must (by the preceding article) be equal to the sum of the two angles EFG and EGF; but by supposition it is equal to the angle EFG alone; therefore the angle AEF must be equal to the sum of the two angles EFG and EGF, and at the same time equal to EFG alone, which is absurd; therefore the lines AB, CD, cannot meet, and must be parallel.

A E B

C

F D

XXXV.

In any right-lined triangle, ABC, the sum of the three angles is equal to two right angles.

To prove this, you must produce BC (in the fig. Art. 33) towards D; then (by Art. 33) the external angle ACD ABC + BAC; to both add the augle ACB, and we have ACD+ACB ABC+BAC+ACB; but ACD+ACB=two right angles (by Art. 27). Hence, ABC+BAC+ACB: two right angles; therefore the sum of the three angles of any plain triangle, ACB, is equal to two right angles.

XXXVI.

Hence in any plain triangle, if one of its angles be known, the sum of the other two will be also known.

For by the last article the sum of all three angles is equal to two right angles, or 180°; hence, by subtracting the given angle from 180°, the remainder will be the sum of the other two.

In any right-angled triangle, the two acute angles taken together are just equal to a right angle; for, all three angles being equal to two right angles, and one angle being right by supposition, the sum of the other two must be equal to a right angle; consequently, any one of the acute angles being given, the other one may be found by subtracting the given one from 90 degrees.

XXXVII.

If in any two triangles, ABC, DEF, two legs of the one, AB, AC, be equal to two legs of the other, DE, DF, each to each respectively, that is, AB=DE, and AC = DF, and the angles BAC, EDF, included between the equal legs be equal; then the remaining leg of the one will be equal to the remaining leg of the other, and the angles opposite to the equal legs will be equal; that is, BC= EF, ABC=DEF, and ACB=DIE.

For if the triangle ABC be supposed to be lifted up and put upon the triangle DEF, with the point A on the point D, and the line AB upon DE, it is plain, since AB=DE, that the point B will fall upon E; and since the angles BAC, EDF are equal, the line AC will fall upon DF; and these lines being of equal length, the point C will fall upon F; consequently the line BC will fall exactly upon the line EF, and the triangle ABC will in all respects be exactly equal to the triangle DEF, and the angle ABC will be equal to the angle DEF, also the angle ACB will be equal to the angle DFE.

XXXVIII.

B

After the same manner it may be proved that if in any two triangles, ABC, DEF (see the preceding figure), two angles, ABC and ACB, of the one be equal to two angles, DEF, DFE, of the other, and the included side, BC, be equal to EF, the remaining sides and included angles will also be equal to each other respectively; that is, AB=DE, AC=DF, and the angle BAC=the angle EDF.

For if the triangle ABC be supposed to be lifted up and laid upon the triangle DEF, the point B being upon the point E, and the line BC upon the line EF, then, since BCEF, the point C will fall upon the point F; and, as the angle ACB the angle DFE, the line CA will fall upon the line FD; by the same way of reasoning, the line BA will fall upon the line ED; therefore the point of intersection, A, of the two lines, BA, CA, will fall upon D, the point of intersection of the lines ED, FD; consequently AB=DE, AC=DF, and the angle BAC= the angle EDF.

XXXIX.

If two sides of a triangle are equal, the angles opposite these sides will also be equal; that is, if ABAC, the angles ABC, ACB, will also be equal.

For, draw the line AD, bisecting the angle BAC, and meeting the line BC in D, dividing the triangle BAC into two triangles, ABD, ACD, in which the side AB AC, the side AD is common to both triangles, and the angle BAD the angle DAC; consequently (by Art. 37), the angle ABD must be equal to the angle ACD.

B

A

C

The converse of this proposition is also true; that is, if two angles of a triangle are equal, the opposite sides are also equal. This is demonstrated nearly in the same

XL.

Any angle at the circumference of a cisle is equal to half the angle at the centre. standing upon the same arc.

To

Thus the angle BAD is half the angle BCD, standing upon the same arc, BD, of the circle BEDA whose centre is Č. demonstrate this, draw through A and the centre C, the right line ACE; then (by Art. 33) the angle CAD+angle CDA = angle ECD; but AC=CD (being two radii of the same circle); therefore

(by Art. 39), the angle CAD the angle CDA, and the sum of these B two angles is the double of either of them; that is, CAD +CDA

=

E

twice CAD; therefore ECD=twice CAD; in the same manner it may be proved that BCE twice BAC, and by adding these together, we have ECD+BCE=twice CAD+twice BAC; that is, BCD=twice BAD, or BAD equal to half of BCD. The demonstration is similar when B, D, fall on the same side of E.

XLI.

An angle at the circumference is measured by half the arc it subtends. For an angle at the centre, standing on the same arc, is measured by the whole arc (by Art. 11); but since an angle at the centre is double that at the circumference (Art. 40), it is evident that an angle at the circumference must be measured by half the are it stands upon. Hence all angles, ACB, ADB, AEB, &c., at the circumference of a circle standing on the same chord, AB, are equal to each other; for they are all measured by the same arc, viz. half the arc AB.

XLII.

An angle in a segment greater than a semicircle is less than a right angle.

Thus, if ABC be a segment greater than a semicircle, the arc AC on which it stands must be less than a semicircle, and the half of it less than a quadrant or a right angle; but the angle ABC in the segment is measured by the half of the arc AC; therefore it is less than a right angle.

An angle in a semicircle is a right angle.

For since DEF is a semicircle, the arc DKF must also be a semicircle; but the angle DEF is measured by half the arc DKF, that is, by half a semicircle or by a quadrant; therefore the angle DEF is a right one.

An angle in a segment less than a semicircle is greater than a right angle.

Thus, if GHI be a segment less than a semi-circle, the arc GKI on which it stands must be greater than a semicircle, and its half greater than a quadrant or right angle; therefore the angle GHI, which is measured by half the arc GKI is greater than a right angle.

XLIII.

If from the centre, C, of the circle ABE there be let fall the perpendicular CD on the chord AB, it will bisect the chord in the point D.

Draw the radii CA, CB; then (by Art. 39) the angle CBA=the angle CAB, and as the angles at D are right, the angle ACD must be equal to the angle BCD (by Art. 36). Hence in the triangles ACD, BCD, we have the angle ACD equal to the angle BCD, CA CB, and CD common to both triangles, consequently (by Art. 37) AD= DB; that is, AB is bisected at D.

XLIV.

E C

D

A

If from the centre, C, of the circle ABE there be drawn a perpendicular CD, to the chord AB, and it be continued to meet the circle in F, it will bisect the are AFB in F (See the preceding figure.)

For in the last article it was proved' that the angle ACD the angle BCD; hence (by Art. 11) the arc AF the arc FB.

=

XLV.

Any line bisecting a chord at right angles is a diameter.

For since by Art. 43) a line drawn from the centre perpendicular to a chord, bisects that chord at right angles, therefore conversely a line bisecting a chord at right angles, must pass through the centre, and consequently be a diameter.

XLVI.

The sine of any arc is equal to half the chord of twice that arc.

For (in the last scheme) AD is the sine of the arc AF, and AF is equal to half the arc AFB, and AD half the chord AB; hence the proposition is manifest.

XLVII.

If two equal and parallel lines, AB, CD, be joined by two others, AC, BD, these will be also equal and parallel.

A

B

To demonstrate this, join the two opposite angles A and D with the line AD; then it is evident, that the line AD divides the quadrilateral ACDB into two triangles, ABD, ACD, in which AB is equal to CD, by supposition, and AD is common to both triangles; and since AB is parallel to CD, the angle BAD is equal to the angle ADC (by Art. 30); therefore, in the two triangles, the sides AB, AD, and the angle BAD, are equal respectively to the sides CD, AD, and the angle ADC; hence (by Art. 37) BD is equal to AC, and the angle DAC equal to the angle ADB; therefore (by Art. 34) the lines BD, AC, must be parallel.

Cor. Hence it follows, that the quadrilateral ABDC is a parallelogram, since the opposite sides are parallel. It is also evident that, in any parallelogram, the line joining the opposite angles (called the diagonal), as AD, divides the figure into two equal parts, since it has been proved that the triangles ABD ACD, are equal to each other.

XLVIII.

It follows also from the preceding article, that a triangle, ACD (see the preceding figure), on the same base, and between the same parallels with a parallelogram, ABDC, is the half of that parallelogram.

XLIX.

From the same article it also follows, that the opposite sides of a parallelogram are equal; for it has been proved, that, ABDC being a parallelogram, AB is equal to CD, and AC equal to BD.

L.

A с

K

E

All parallelograms on the same or equal bases, and between the same parallels, are equai to each other; that is, if BD and GH be equal, and the lines BH, ÁF, be parallel, the parallelograms ABDC, BDFE, and EFHG, will be equal to each other. For AC is equal to EF, each being equal to BD (by Art. 49); to both add CE, and we have AE, equal to CF; therefore in the two triangles ABE, CDF, AB is equal to CD, AE is equal to CF, and the angle BAE is equal to DCF (by Art. 31); therefore the two triangles ABE, CDF, are equal (by Art. 37), and taking the triangle CKE from both, the figure ABKC is equal to the figure KDFE, to both which add the triangle KBD, and we have the parallelogram ABDC, equal to the parallelogram BDFE. In the same way it may be proved that the parallelogram EFHG is equal to the parallelogram BDFE; therefore the three parallelograms ABDC, BDFE, and EFHG, are equal to each other.

B D

G H

Cor. Hence it follows, that triangles on the same base, and between the same parallels, Tre equal, since they are the half of the parallelograms on the same base and between the same parallels (by Art. 48).

LI.

In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the two sides. Thus, if BAC be a right-angled triangle, the square of the

hypotenuse BC, viz. BCMH, is equal to the sum of the squares-made on the two sides, AB and AC, viz. to ABDE and ACGF.

To demonstrate this, through the point A draw AKL perpendicular to the hypotenuse

D

E

and BH equal to BC; therefore in the triangles DBC, ABHI, the two legs, DB, BC, of the one, are equal to the two legs, AB, BH, of the other; and the included angles, DBC and ABH, are also equal; (for DBA is equal to CBH, being both right; to each add ABC, and we have DBC, equal to ABH); therefore the triangles DBC, ABH, are equal (by Art. 37); but the triangle DBC is half of the square ABDE (by Art. 48), and the triangle ABH is half the parallelogram BKLH (by the same article); consequently the square ABDE is equal to the parallelogram BKLÍ. In the same way it may be proved that the square ACGF is equal to the parallelogram KCML. Therefore the sum of the squares ABDE and ACGF is equal to the sum of the parallelograms BKLH and KCML; but the sum of these parallelograms is equal to the square BCMH; therefore the sum of the squares on AB and AC is equal to the square on BC.

B K

H

M

L

Cor. Hence, in any right-angled triangle, if we have the hypotenuse and one of the legs, we may easily find the other leg, by taking the square of the given leg from the square of the hypotenuse; the square root of the remainder will be the sought leg. Thus, if the hypotenuse was 13, and one leg was 5, the other leg would be 12, for the square of 5 is 25, and the square of 13 is 169; subtracting 25 from 169 leaves 144, the square root of which is 12. If both legs are given, the hypotenuse may also be found by extracting the square root of the sum of the squares of the legs; thus, if one leg was 6, and the other 8, the square of the first is 36, the square of the second is 64; adding 36 and 64 together gives 100, whose square root is 10, which is the sought hypotenuse.

LII.

Four quantities are said to be proportional, when the magnitude of the first compared with the second is the same as the magnitude of the third compared with the fourth.

Thus 4, 8, 12, and 24, are proportional, because 4 is half of 8, and 12 is half of 24; and if we take equi-multiples, Aa, A× b, of the quantities a and b, and other equi-multiples, BXa, BX b, of the same quantities a and b, the four quantities, Axa, Ab, Ba, Bb, will be proportional; for AX a compared with Ab is of the same magnitude as a compared with b, and BX a compared with Bb is also of the same magnitude as a compared with b.

LIII.

In any triangle, AGg, if a line, Ee, be drawn parallel to either of the sides, as Gg, the side AG will be to AE as Ag to Ae, or as Gg to Ee.

N

F

H

K

M

e

To demonstrate this, upon the line AG take the line AB so that a certain multiple of it may be equal to AE, and another inultiple of it may be equal to AG; this may be always done accurately when AE and AĞ are commensurable; if they are not accurately commensurable, the quantity AB may be taken so small that certain multiples of it may differ from AE and AG respectively by quantities less than any assignable. On the line AG, take BC, CD, DE, EF, FG, &c., each equal to AB; and through these points draw the lines Bb, Cc, &c., parallel to Gg, cutting the line Ag in the points b, c, d, e, &c.; draw also the lines BM, CL, DK, &c., parallel to Ag, cutting the former parallels in the points N, O, P, &c., and the line Gg in the points M, L, K, &c. Then the triangles ABb, BCN, CDO, &c., are similar and equal to each other; for the lines Bb, CN, are parallel; therefore the angle ABb=BCN (by Art. 31), and by the sane article the angle BAb is equal to CBN (because BN is parallel to Ab), and by construction AB=BC; therefore (by Art. 38) the triangles ABb and BCN are equ to each other; and in the same manner we may prove that the others, CDO, DEP, EFQ, &c., are equal to ABb. Therefore Ab=BN=CO=DP, &c., and Bb CN=DO= EP, &c. ; but (by Art. 49) BN=bc, CO=cd, DP=de; therefore Abbccdde, &c.; and since (by construction) AB=BC=CD, &c., any AE is the same multiple of AB as the corresponding line Ae is of Ab; and AG is the same multiple of AB as Ag is of Ab; therefore the lines AG, AE, Ag, Ae, are

line