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Exercise 1.-In a triangle, the straight line drawn from the vertex bisecting the base, bisects every parallel to the base intercepted by the sides. Exercise 2.-In equiangular triangles, the perpendiculars drawn from the vertices of equal angles to the opposite sides, are proportional to those sides.

PROP. V. THEOREM.

If the sides of two triangles, about each of their angles, be proportionals, the triangles are equiangular; and the equal angles are those which are opposite to the homologous sides.

Let the triangles ABC and DEF have their sides proportionals, so that A B is to BC, as D E to EF; and B C is to CA, as EF to FD; and therefore, ex æquali, B A is to A C, as ED is to DF. The triangle ABC is equiangular to the triangle DEF, and the angles which are opposite to the homologous sides are equal, viz. the angle ABC equal to the angle DEF, the angle BCA to the angle EFD, and the angle BAC to the angle EDF.

At the points E and F, in the straight line EF, make the angle FEG equal to the angle ABC, and the angle EF G equal to BCA (I. 23).

Because the remaining angle EG F, is equal to the remaining angle BAC (I. 32), the triangle GEF is equiangular to the triangle ABC. There

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fore they have their sides opposite to the equal angles proportionals (VI. 4). Wherefore, A B is to BC as GE is to EF. But A B is to B C. as DE is to EF (Hyp.). Therefore D E is to EF, as GE is to EF (V. 11). Because D E and G E have the same ratio to EF, DE is equal to GE (V. 9). For the same reason, DF is equal to F G. Because, in the triangles DEF and GEF, DE is equal to E G, and E F is common, the two sides DE and EF are equal to the two GE and EF, each to each. But the base D F is equal to the base G F. Therefore the angle DEF is equal to the angle GEF (I. 8), and the remaining angles of the one to the remaining angles of the other, each to each. Therefore the angle DFE is equal to the angle GFE, and the angle EDF to the angle EGF. Because the angle DEF is equal to the angle G EF, and the angle GEF is equal to the angle ABC (Const.). Therefore the angle A B C is equal to the angle DEF (I. 4x. 1). For the same reason, the angle A CB is equal to the angle D FE, and the angle at A is equal to the angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q. E. D.

This proposition is the converse of Prop. IV. The 4th and 5th of Book VI., and the 47th and 48th of Book I., constitute the most important principles in the Elements. They include the principles of Trigonometry and its various applications, as well as those of analytic geometry in general.

PROP. VI. THEOREM.

If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals; the triangles are equiangular, and those angles are equal which are opposite to the homologous sides.

Let the triangles ABC and DEF have the angle BAC in the one.

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equal to the angle EDF, in the other; and the sides about those angles proportionals, that is, BA to AC, as ED to DF. The triangles ABC and DEF are equiangular, and the angle ABC is equal to the angle DEF, and the angle ACB to the angle DFE.

At the points D and F, in the straight line DF, make the angle FDG equal to either of the angles BAC or EDF (I. 23); and the angle DFG ecual to the angle ACB.

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Because the remaining angle at B is equal to the remaining angle at G (I. 32), the triangle B DGF is equiangular to the triangle. ABC. Therefore BA is to AC, as GD is to DF (VI. 4). But (Hyp.) BA is to AC, as ED is to DF. Therefore ED is to DF, as GD is to DF (V. 11); and ED is equal to DG (V. 9). Because ED is equal to DG and DF is common to the two triangles EDF and GDF. Therefore the two sides ED and DF are equal to the two sides GD and DF, each to each. But the angle EDF is equal to the angle GDF (Const.). Therefore the base EF is equal to the base FG (I. 4), and the triangle EDF to the triangle GDF. Because the remaining angles of the one are equal to the remaining angles of the other, each to each. Therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E. But the angle DFG is equal to the angle A CB (Const.). Therefore the angle ACB is equal to the angle DFE (I. Ax. 1). But the angle BAC is equal to the angle EDF (Hyp.). Therefore the remaining angle at B is equal to the remaining angle at E (I. 32); and the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D.

Corollary 1.-Triangles which have a common angle and the sides about it proportionals, have their bases parallel, and are equiangular and similar.

Corollary 2.-Triangles which have a common angle and parallel bases, are equíangular and similar.

Exercise. If from any number of points in a straight line parallels be drawn proportional to the distances or these points from a given point, the straight line joining this point and the extremity of one of the parallels passes through the extremities of all the parallels.

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If two triangles have one angle of the cne equal to one angle of the other, and the sides about another angle in each, proportionals; and if the remaining angle in each be of the same affection (that is, either both acute, or both not acute); the two triangles are equiangular and similar.

Let the two triangles A B C and DEF have one angle BAC in the one equal to the angle EDF in the other, and the sides about their two other angles ABC and DEF, proportionals, so that A B is to BC as DE is to EF; and let their remaining angles ACB and DFE be of the same affection, that is, either both acute, or both not acute. The triangles A B C and D E F are equiangular and similar.

For if the angle ABC is not equal to the angle DEF one of them must be greater than the other. Let ABC be the greater angle, and at the point B in the straight line A B, make the angle ABG equal to the angle D E F (I. 23).

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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 Bis 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.

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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 B C. The triangles ABD and ADC are similar to the whole triangle A B C, and to one another.

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Because the angle BAC is equal to the angle ADB, 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 ABD and ACD, are both

equiangular and similar to ABC. 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 B A, as BA is to BD; and in the triangles A B C and ACD, B C is to CA, as CA 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 BA C, 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.

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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.

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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 BC. 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 Fand G, into parts proportional to those of A C. Through D draw DHK parallel to A B (I. 31).

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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 K H 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.

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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.

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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 equal to AC. Therefore AB is to AC, as AC is to CE (V. 7). Wherefore, to the two given straight lines AB and AC, a third proportional CE is found. Q. Ë. 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 B C, 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.

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