describe a circle cutting Ab in E; then with centre E and the same radius as before describe a circle cutting Ab in C; so that AD = DE = EC. Join CB, and through E and D draw EG and DF parallel to BC, cutting AB in G and F. Then since AD = DE= EC, and CB, EG, DF are parallels, .. also AF = FG = GB, that is, AB is divided into three equal parts in the points F and G, (67 Cor. 1.)

69. PROP. III. If any two straight lines be cut by three parallel straight lines, the parts intercepted between the parallels shall be 'proportionals'.

Let ABC, DEF be any two straight lines, and AD, BE, CF, three parallel straight lines intersecting the former in A, B, C, and D, E, F. Then AB, BC, DE, EF shall be proportionals', that is, AB shall have the same ratio to BC that DE has to EF.

Let AB contain the line which is the unit of measurement three times, and BC the same unit five

B

b

times, (the same proof will hold for any numbers), and divide AB into three equal parts, and BC into five (68), and through the points of division draw lines parallel to BE intersecting DE, and EF: then DE will be divided by these parallels into the same number of equal parts as AB, and EF into the same number as BC (67), that is, DE will contain a certain line three times, and EF the same line five times; or the ratio of DE to EF is three to five, which is the same ratio as AB to BC; .. AB, BC, DE, EF are proportionals (66).

Observe, it may be that a specified unit of measurement will not exactly divide AB, and BC, in which case the unit must be reduced, until this can take place. For example, if there be not an exact number of feet in AB, and BC, there may be an exact number of inches; or, if not inches, there may be an exact number of tenths of an inch; and so on. And whatever be the reduced unit which will exactly divide both AB and BC, the proof above given then holds.

70. PROP. IV. If two sides of a triangle be intersected by a straight line parallel to the third side, the two sides are divided proportionally.

Taking the preceding fig. in (69), through A draw Abc parallel to DEF, cutting BE in b, and CF in c. Then ACc will represent any triangle having its two sides AC, Ac, intersected in B, b, by Bb which is parallel to Cc; and it is required to prove that AB is to BC as Ab is to bc.

=

Simi

Since ADEb is a parallelogram, Ab = DE. larly be EF; and, by (69), it is proved that AB is to BC as DE is to EF, .. AB is to BC as Ab is to be that is, AC, Ac are divided proportionally in B, b.

COR. It follows that AB is to AC as Ab is to Ac. For since AB is to BC as Ab is to bc, this means that AB contains, or is contained in, BC, the same number of times that Ab contains, or is contained in, bc. Now it is plain that, whatever be the number of times AB contains, or is contained in, BC, it is contained in, AB+ BC, or AC, exactly once more. Also whatever be the number of times Ab contains, or is contained in, bc, it is contained in, 4b+bc, or Ac, exactly once more. But, if each of two equal numbers be increased by 1, they will remain equal; AB is contained in AC the same number of times that Ab is in Ac; that is, the ratio of AB to AC is equal to the ratio of Ab to Ac.

The common mode of writing the two last results is,
AB: BC :: Ab: bc, and AB: AC :: Ab: A c.

71. PROP. V. Similar triangles have the sides forming the equal angles proportionals.

[DEF. Similar triangles are such as have their angles equal, each to each.]

Let ABC, abc, be similar triangles, that is, A= Lα,

< B=2b, and <C= c; it is required to shew that

L

ab: AB :: ac: AC,

ab: AB:: bc: BC,

and bc BC: ac: AC.

:

With centre A and radius ab describe a circle cutting AB in D, and another circle with the same centre and radius ac cutting AC in E, and join ED. Then in the two triangles ADE, abc, the two sides AD, AE, are equal to the two sides ab, ac, each to each, and ‹DAE +bac,.. the triangles are equal in all respects (24), and .. 4 ADE=abc. But abc = ‹ ABC, :. LADE = < ABC, and_.. DE is parallel to BC (34). Hence by (70 Cor.) AD : AB :: AE : AC; but AD = ab, and AE= ac,

. ab AB :: ac: AC.

:

In the same way, by making B the centre of the circles, it may be shewn, that

ab AB: bc : BC;

and by making C the centre, that

bc BC: ac: AC.

COR. 1. Conversely, if two triangles have an angle of one equal to an angle of the other, and the sides forming the equal angles proportionals, the triangles will be similar.

COR. 2. Hence, also, if a triangle be cut off from a larger triangle by a line parallel to one of the sides, the two triangles will be similar, and have the sides about equal angles proportionals.

COR. 3. If CD be drawn perpendicular to AB, and cd to ab, it will also easily appear, that

CD: cd: AC: ac, or :: AB: ab, or :: BC: bc.

72. PROP. VI. If a right-angled triangle be divided into two other right-angled triangles by a straight line drawn from the vertex of the right angle perpendicular to the opposite side, each of these two triangles shall be similar to the whole triangle and to one another.

Let ABC be a right-angled triangle having ‹ BAC the right angle. From A draw AD perpendicular to BC; then the triangles ABD, ADC shall be similar to the triangle ABC, and to each other.

Because ‹ BAC=a_right angle =< ADB, and B is

common to the two triangles ADB, ABC; and since the three

angles of every triangle are to

gether equal to two right angles, (37), the remaining 4 BAD of the one triangle is equal to the

B D

remaining ACB of the other, that is, the triangles ADB, ABC, are equiangular, and .. similar.

In the same way it may be shewn, that the triangles ADC, ABC, are equiangular, and .. similar. Hence also the triangles ADB, ADC are equiangular and similar.

COR. 1. Since in similar triangles the sides forming equal angles are proportionals (71); and since the triangles ADB, ADC are similar, .. BD : AD :: AD : DC. Again, since ABD and ABC are similar triangles, BD: AB :: AB : BC.

Also, since ADC and ABC are similar,

CD: AC AC: BC.

[DEF. When of four magnitudes which are proportionals the second and third are the same, this latter magnitude is said to be a mean proportional between the other two.

Thus, in this Cor. AD is a mean proportional between BD and DC. Also AB is a mean proportional between BD and BC; and AC is a mean proportional between BC and CD.]

D

COR. 2. Hence to find a mean proportional between two given straight lines, AB, BC, place AB, BC so as to form one straight line AC. Upon AC describe a semicircle; through B draw BD at right angles to AC meeting the circumference in D; and join AD, CD. Then, since ▲ ADC is a right A

B

angle, and DB is perpendicular to AC, by Cor. 1, AB : BD:: BD: BC; :. BD is a mean proportional between AB and BC.

73. PROP. VII. The areas of parallelograms or triangles between the same parallels are proportional to their bases.

D C

(1) Let ABCD, BEFC be two parallelograms between the same parallels ABE, DCF, and upon the bases AB, BE, respectively.

Let such a lineal unit of measurement be taken as will exactly divide both AB, and

B

BE; and divide AB and BE ▲ into as many equal parts as they contain the unit (68). Through the several points of division draw lines parallel to AD or BC, dividing ABCD into as many parallelograms as the unit is contained in AB, and BEFC into as many as the unit is contained in BE. Then since parallelograms upon equal bases and between the same parallels are equal to one another (41 Cor. 1), the smaller parallelograms which make up ABCD, and BEFC, are all equal. Therefore the ratio of ABCD to BEFC will be the ratio of the sum of these equal parallelograms in the one to the sum of them in the other, that is, as the number of them in the one to the number in the other, (since they are all equal) or as the number of units in AB is to the number of units in BE, that is, as AB is to BE.

(2) Again, since a parallelogram is double of the triangle upon the same base and between the same parallels (40), and the halves of two magnitudes will plainly bear the same ratio to each other that the whole magnitudes do,.. joining BD, EC, the triangle ABD, which is half of the parallelogram ABCD, will have the same ratio to the triangle BEC, which is half of BEFC, that AB has to BE.

COR. It follows also, that the areas of triangles or parallelograms of equal altitudes, however situated, are proportional to their bases; the altitude being the perpendicular let fall from the vertex of one of the angles upon the opposite side considered as the base.

74. PROP. VIII. If four straight lines taken in order be proportionals', the rectangle contained by the