A

G

CE

A

D

Because in the two triangles ABG and DEF, the angle at A is equal to the angle at D (Hyp.) and the angle ABG is equal to the angle DEF (Const.). Therefore the remaining angle AGB is equal to the remaining angle DFE (I. 32), and the triangle ABG is equiangular to DEF. Because these triangles are equiangular AB is to BG, as DE is to EF (VI. 4). But A B is to BC as DE is to EF (Hyp.). Therefore B AB is to BC, as AB is to BG (V. 11); and BG is equal to BC (V. 9). Wherefore the triangle GBC is isosceles, and the angle BGC is equal to the angle BCG. Therefore the angles BGC and BGA are both acute, or both not acute, according as the angles ACB and DFE, are both acute or both not acute. If they are both acute, as in the first figure, the angles BGC and BGA which BG makes with AC are together less than two right angles, which is impossible (I. 13). If they are both not acute, as in the second figure, the angles BGC and BCG of the triangle BGC, are together not less than two right angles, which is also impossible (I. 17). Therefore the angle ABC is not unequal to the angle DEF; that is, the angle ABC is equal to the angle DEF. Wherefore the remaining angle ACB is equal to the remaining angle DFE; and the triangle ABC is equiangular and similar to the triangle DEF. Q. E. D.

B

CE

F

In this proposition, we have considerably departed from Euclid, in order to make
the demonstration more plain and easy to the learner. Besides, Euclid has
three separate cases to demonstrate, which are here reduced to two, and the
same proof is made, with a very slight exception, to apply to both.
Corollary.-If two triangles have two sides in the one proportional to two sides in
the other, and the angles opposite one pair of the homologous sides equal,
the angles opposite the other pair of homologous sides are either equal or
supplementary.

PROP. VIII. THEOREM.

In a right-angled triangle, if a perpendicular be drawn from the right angle to the opposite side, the triangles on each side of the perpendicular are similar to the whole triangle, and to one another.

Let A B C be a right-angled triangle, having the right angle BAC, and from the point A let AD be drawn perpendicular to the opposite side BC. The triangles ABD and ADC are similar to the whole triangle A B C, and to one another.

B

A

D C

Because the angle BAC is equal to the angle AD B, each of them being a right angle (I. Ax. 11), and the angle at B is common to the two triangles ABC and ABD. Therefore the remaining angle ACB is equal (I. 32) to the remaining angle B AD; and the triangle A B C is equiangular to the triangle ABD. But the sides about their equal angles are proportionals (VI. 4). Therefore the triangles are similar (VI. Def. 1). In like manner it may be shown that the triangle ADC is equiangular and similar to the triangle ABC. Because the triangles A B D and ACD, are both

K 2

equiangular and similar to AB C. Therefore they are equiangular and similar to each other. Therefore, in a right-angled, &c. Q. E. D.

COROLLARY.-From this it is manifest, that the perpendicular drawn from the right angle of a right-angled triangle to the hypotenuse, is a mean proportional between the segments of the hypotenuse; and also that each of the sides is a mean proportional between the hypotenuse, and the segment of it adjacent to that side. For in the triangles BDA and ADC, B D is to DA, as DA is to D C. In the triangles ABC and DBA, BC is to BA, as BA is to BD; and in the triangles ABC and ACD, B C is to CA, as C A is to CD (VI. 4).

The preceding corollary to this proposition is the most important part of it, as will be seen by its subsequent application. From this corollary, in combination with Props. XVI. and XVII. of this Book, and Prop. II. Book II., a new and independent demonstration of Prop. XLVII. Book I., may be obtained, which is considerably shorter than that given by Euclid.

Corollary 1.-The hypotenuse is to either leg of the right-angled triangle, as the other leg is to the perpendicular drawn to the hypotenuse from the right angle. Corollary 2.-In a triangle, if the triangles formed by the perpendicular and the sides be similar, the angles which it makes with the sides must be either equal or complementary. If the angles be equal, the perpendicular bisects the vertical angle, the similar triangles are equal, and the whole triangle is isosceles. If these angles be not equal, and the perpendicular be within the triangle, they must be equal to the alternate angles at the base, complementary to each other, and the vertical angle being then equal to the sum of the angles at the base, is a right angle. If the perpendicular be without the triangle, the angles between it and the sides are complementary, and the vertical angle being equal to the difference between these angles, is an acute angle.

Corollary. If the base, the two sides, and the perpendicular of a triangle be proportionals, it is a right-angled triangle.

PROP. IX. PROBLEM.

From a given straight line to cut off any required part, or submultiple. Let AB be the given straight line. It is required to cut off any part or submultiple from it.

From the point A draw a straight line A C, making any A angle BAC, with A B. In AC take any point D, and make AC the same multiple of AD, that A B is of the part to be cut off from it. Join B C, and draw DE parallel to CB. The part AE is the part required to be cut off from A B.

E

B

D

Because ED is parallel to BC, one of the sides of the triangle ABC, CD is to DA, as BE is to EA (VI. 2). Therefore by composition, CA is to AD, as BA is to AE (V. 18). But CA is a multiple of AD (Const.). Therefore B A is the same multiple of AE (V. D); and whatever part AD is of A C, AE is the same part of AB. Wherefore, from the straight line AB is cut off the part or submultiple required. Q. E. F.

Corollary. By drawing through the points in A C, which mark off the successive parts of it each equal to AD, parallels to the straight line BC the line AB will be divided into the same number of equal parts. Thus a given straight line may be divided into any number of equal parts required.

To divide a given straight line similarly to a given divided straight line, that is, into parts proportional to the parts of the given divided straight line.

Let A B be the given straight line to be divided, and AC the given straight line, divided into parts at D and E. It is required to divide AB into parts proportional to the parts of A C.

Place A B and A C so as to make any angle BAC with each other; and join B C. Through the points D and E draw DF and E G parallels to B C (I. 31). The straight line A B is divided, at the points F and G, into parts proportional to those of A C. Through D draw DHK parallel to A B (I. 31).

A

F

E

G

B

K

C

Because HE

Because FH and HB are parallelograms (Const.). Therefore D H is equal to F G, and HK to GB (I. 34). is parallel to K C, one of the sides of the triangle DKC, CE is to ED, as KH is to HD (VI. 2). But BG is equal to KH, and GF to HD (Const.). Therefore C E is to ED, as BG is to GF (V. 7). Again, because FD is parallel to G E, one of the sides of the triangle AGE, ED is to DA, as GF is to FA (VI. 2). Therefore, as has been proved, CE is to ED, as BG is to GF, and ED is to DA, as GF is to FA. Wherefore the given straight line AB is divided into parts proportional to the parts of AC. Q. E. F.

Exercise 1.-To divide a given straight line, into two parts that shall have to each other a given ratio.

Exercise 2.-To produce a given straight line, so that the whole line thus produced may have to the part produced a given ratio.

To find a third proportional to two given straight lines.

Let AB and AC be the two given straight lines. It is required to find a third proportional to AB and AC.

Place AB and AC so as to make any angle BAC with each other. Produce AB and AC to the points D and E; making BD equal to AC (I. 3). Join BC, and through D, draw DE parallel to BC (I. 31). The straight line CE is a third proportional to AB and A C.

B

Because BC is parallel to DE, a side of the triangle ADE, AB is to BD, as AC is to CE (VI. 2). But BD is D equal to AC. Therefore AB is to AC, as AC is to CE

A

E

(V. 7). Wherefore, to the two given straight lines AB and AC, a third proportional CE is found. Q. E. F.

Exercise. If in the preceding figure, instead of making BD equal to AC, AD had been made equal to A C, and D E drawn parallel to BC, then AE would have been the third proportional. Required the demonstration.

This problem may be solved in a variety of ways, especially by the application of Prop. VIII., and its corollary. It will be useful exercise for the student, to find out some of these solutions. In the same way, the problem may be extended to the mode of finding a series of continual proportionals.

PROP. XII. PROBLEM.

To find a fourth proportional to three given straight lines.

Let A, B, and C be the three given straight lines. It is required to find a fourth proportional to A, B, and C.

Take two straight lines DE and DF, containing any any angle EDF. Upon these straight lines make DG equal to A, GE equal to B, and DH equal to C (1.3). Join GH, and through E draw EF parallel to GH (I. 31). The straight line HF is the fourth proportional to A, B, and Č.

Because G H is parallel to EF, one of the sides of the triangle DEF, DG is to GE, as DH is to HF E (VI. 2). But DG is equal to A, GE to B, and DH to

C.

D

H

F

A

B

C

Therefore A is to B, as C is to HF (V. 7). Wherefore, to the three given straight lines A, B, and C, a fourth proportional HF is found. Q. E. F.

This problem may also be solved in a variety of ways, chiefly by the application of Props. XXXV. and XXXVI. Book III., and Prop. VIII. of this Book. The ingenious student will find it a useful exercise to try and find out these solutions.

To find a mean proportional between two given straight lines.

Let AB, and BC be the two given straight lines. It is required to find a mean proportional between them.

Place AB and BC in a straight line adjacent to each other; and upon AC describe the semi-circle ADC. From the point B, draw BD at right angles to AC (I. 11). The straight line BD is a mean proportional between AB and BC. Join AD and DC.

D

B

C

There

Because the angle ADC in a semicircle is a right A (III. 31), and BD is drawn from the right angle perpendicular to the opposite side AC of the triangle ADC. fore DB is a mean proportional between AB and B C, the segments of the base (VI. 8, Cor.). Wherefore between the two given straight lines A B and B C, a mean proportional D B is found. Q. E. F.

Other modes of solving this problem are suggested to the student by the consideration of Prop. VIII. of this Book, and Props. XXXI. and XXXVI, of Book III. To find two mean proportionals between two given straight lines, is beyond the power of elementary geometry. This problem was connected with the famous problem of the "duplication (doubling) of the cube;" for, if four straight lines be continually proportional, the first is to the fourth, in the triplicate ratio of the first to the second; or, as the cube of the first is to the cube of the second.

PROP. XIV. THEOREM.

Equal parallelograms, which have an angle of the one equal to an angle of the other, have their sides about the equal angles reciprocally proportional; and conversely, parallelograms which have an angle of the one equal to an angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

First, Let AB and BC be equal parallelograms, which have their

angles at B equal. The sides of the parallelograms A B and B C about the equal angles, are reciprocally proportional; that is, DB is to BE, as GB is to BF.

Place the parallelograms so that the sides DB and BE shall be in the same straight line; FB and B G shall be in one straight line (I. 14). Complete the parallelogram F E.

A

F

E

D

B

G C

Because the parallelogram AB is equal to the parallelogram BC, and FE is another parallelogram. Therefore A B is to FE, as BC is to FE (V. 7). But AB is to FE, as the base DB is to the base BE (VI. 1); and BC is to FE, as the base GB is to the base B F. Therefore DB is to BE, as GB is to BF (V. 11). Wherefore the sides of the parallelograms AB and B & about their equal angles are reciprocally proportional.

Next, let the sides about the equal angles of the parallelograms A B and BC be reciprocally proportional,-viz., DB to B E, as G B to B F. The parallelogram A B is equal to the parallelogram B C.

Because, D B is to BE, as GB is to B F. But DB is to B E, as the parallelogram A B is to the parallelogram FE (VI. 1); and G B is to BF as the parallelogram B C is to the parallelogram FE. Therefore AB is to FE, as BC is to FE (V. 11). ́ Wherefore the parallelogram AB is equal to the parallelogram BC (V. 9). Therefore equal parallelograms, &c. Q. E. D.

Corollary.-If two parallelograms are equal, and have their sides reciprocally proportional, they are equiangular.

Equal triangles which have an angle of the one equal to an angle of the other, have their sides about the equal angles reciprocally proportional; and conversely, triangles which have an angle in the one equa to an ang e in the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

First, let A B C and ADE be equal triangles, which have the angle BAC equal to the angle DAE. The sides about the equal angles of the triangles are reciprocally proportional; that is, CA is to AD, as EA is to AB.

Place the triangles so that their sides CA and AD shall be in one straight line; EA and AB shall be in one straight line (I. 14). Join BD.

B

D

E

Because the triangle ABC is equal to the triangle ADE, and ABD is another triangle. Therefore the triangle CA B, is to the triangle BAD, as the triangle A ED is to the triangle DAB (V. 7). But the triangle CAB is to the triangle BAD, as the base C A is to the base A D (VI. 1); and the triangle EAD is to the triangle DAB as the base È A is to the base AB (VI. 1). Therefore CA is to AD, as EA is to AB (V. 11). Wherefore the sides of the triangles A B C, ADE, about the equal angles are reciprocally proportional.

Next, let the sides of the triangles A B C and ADE about their equal angles at A be reciprocally proportional,-viz., CA to AD, as EA to A B. The triangle ABC is equal to the triangle ADE. Join BD.