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

Demonstration. Because the remaining angle EGF, is equal to the remaining angle BAC

32),

therefore

B

E

1. The triangle GEF is equiangular to the triangle ABC; consequently they have their sides opposite to the equal angles proportional (VI. 4); wherefore

2. As AB to BC, so is GE to EF;

but as AB to BC, so is DE to EF (hyp.); therefore as DE to EF, so is GE to EF (V. 11); that is,

3. DE and GE have the same ratio to EF, and consequently

are equal (V. 9).

For the same reason, DF is equal to FG; and because, in the triangles DEF, GEF; DE is equal to EG, and EF is common; the two sides DE, EF are equal to the two GE, EF, each to each; and the base DF is equal to the base GF; therefore

4. The angle DEF is equal to the angle GEF (I. 8); and the other angles to the other angles which are subtended by the equal sides (I. 4); therefore

5. The angle DFE is equal to the angle GFF, and EDF to EGF.

And because the angle DEF is equal to the angle GEF, and GEF equal to the angle ABC (constr.); therefore

6. The angle ABC is equal to the angle DEF (Ax. 1) ; for the same reason, the angle ACB is equal to the angle DFE, and the angle at A equal to the angle at D; therefore

7. The triangle ABC is equiangular to the triangle DEF.

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PROPOSITION 6.-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 shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.

Let the triangles ABC, DEF have the angle BAC in the one equal to the angle DEF in the other, and the sides about those angles proportionals; that is, BA to AC, as ED to DF.

Then the triangles ABC, DEF shall be equiangular, and shall have the angle ABC equal to the angle DEF, and ACB to DFE.

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

Demonstration. Because the remaining angle at B is equal to the remaining angle at G (I. 32); therefore

1. The triangle DGF is equiangular to the triangle ABC;

and therefore

2. As BA to AC, so is GD to DF (VI. 4);

but, by the hypothesis, as BA to AC, so is ED to DF; therefore as ED to DF, so is GD to DF (V. 11); wherefore

3. ED is equal to DG (V. 9);

and DF is common to the two triangles EDF, GDF; therefore the two sides ED, DF, are equal to the two sides GD, DF, each to each; and the angle EDF is equal to the angle GDF (constr.); wherefore

4. The base EF is equal to the base FG (I. 4), and the triangle EDF to the triangle GDF ;

and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides; therefore

5. 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 ACB (constr.); therefore

6. The angle ACB is equal to the angle DFE (Ax. 1) ;

and the angle BAC is equal to the angle EDF (hyp.); wherefore also

therefore

7. The remaining angle at B is equal to the remaining angle at E (I. 32);

8. The triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q.E.D

PROPOSITION 7.-Theorem.

If tavo triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals; then, if each of the remaining angles be either less, or not less, than a right angle, or if one of them be a right angle; the triangles shall be equiangular, and shall have those angles equal about which the sides are proportionals.

Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz., the angle BAC to the angle EDF, and the sides about two other angles ABC, DEF proportionals, so that AB is to BC, as DE to EF

First. Let each of the remaining angles at C, F, be less than a right angle.

Then the triangle ABC shall be equiangular to the triangle DEF, viz., the angle ABC shall be equal to the angle DEF, and the remaining angle at C equal to the remaining angle at F.

Α

D

4.A

E

F

Construction. For if the angles ABC, DEF be not equal, one of them must be greater than the other; let ABC be the greater, and at the point B, in the straight line AB, make the angle ABG equal to the angle DEF (I. 23).

Demonstration. Then, because the angle at A is equal to the angle at D (hyp.), and the angle ABG to the angle DEF; the remaining angle AGB is equal to the remaining angle DFE (I. 32); therefore

1. The triangle ABG is equiangular to the triangle DEF; wherefore, as AB is to BG, so is DE to EF (VI. 4); but as DE to EF, so, by hypothesis, is AB to BC; therefore

2. As AB to BC, so is AB to BG (V. 11) ;

and because AB has the same ratio to each of the lines BC, BG; BC is equal to BG (V. 9); and therefore

3. The angle BGC is equal to the angle BCG (I. 5) ;

but the angle BCG is, by hypothesis, less than a right angle; therefore also the angle BGC is less than a right angle; and therefore 4. The adjacent angle AGB must be greater than a right angle (I. 13);

but it was proved that the angle AGB is equal to the angle at F; therefore

5. The angle at F is greater than a right angle;

but, by the hypothesis, it is less than a right angle; which is absurd. Therefore

6. The angles ABC, DEF are not unequal, that is, they are equal;

and the angle at A is equal to the angle at D (hyp.); wherefore⚫ 7. The remaining angle at C is equal to the remaining angle at F (I. 32);

therefore

8. The triangle ABC is equiangular to the triangle DEF. Next. Let each of the angles at C, F, be not less than a right angle. Then the triangle ABC shall also in this case be equiangular to the triangle DEF.

B

E

F

Demonstration. The same construction being made, it may be proved in like manner that BC is equal to BG, and therefore

1. The angle at C is equal to the angle BGC;

but the angle at C is not less than a right angle (hyp.); therefore the angle BGC is not less than a right angle; wherefore

2. Two angles of the triangle BGC are together not less than two right angles;

which is impossible (I. 17); and therefore

3. The triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case.

Lastly. Let one of the angles at C, F, viz., the angle at C, be a right angle; in this case likewise, the triangle ABC shall be equiangular to the triangle DEF.

B

C

44

CE

Construction. For, if they be not equiangular, at the point B in. the straight line AB make the angle ABG equal to the angle DEF. Demonstration. Then it may be proved, as in the first case, that BG is equal to BC; and therefore

1. The angle BCG is equal to the angle BGC (I. 5) ; but the angle BCG is a right angle (hyp.); therefore the angle BGC is also a right angle (Ax. 1); whence

2. Two of the angles of the triangle BGC are togeiher not less than two right angles;

which is impossible (I. 17); therefore

3. The triangle ABC is equiangular to the triangle DEF.

Wherefore, if two triangles, &c. Q.E.D.

PROPOSITION 8.-Theorem.

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

Let ABC be a right-angled triangle, having the right angle BAC, and from the point A, let AD be drawn perpendicular to the base BC.

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