PROPOSITION VIII. THEOREM

If two triangles have the three sides of one triangle equal to the three sides of the other, each to oach, they are equal in all respects.

Let ABC, DEF be two triangles having the side AB equal to DE, and AC to DF, and BC to EF; the triangle ABC is equal to DEF, the angle A to D, the angle B to È, and the angle C to F.

Let the triangle DEF be applied to the triangle ABC, so that their longest sides, BC, EF, may coincide; and let BGC represent the triangle DEF in an inverted position. Join AG. Because the sides GB and AB are each equal, by hipothesis, to DE, they are equal to each other; therefore the triangle ABG is isosceles, wherefore the angle BAG is equal to BGA (5. 1.).

In the same manner it may be shown that the side AC is equal to CG, and the angle CAG to CGA. Therefore the two angles BAG, CAG together are equal to the two angles BGA, CGA together; that is, the whole angle BAC is equal to the whole angle BGC. But the angle BGC is, by hypothesis, equal to the angle EDF; therefore also the angle BAC is equal to EDF. Hence the triangles ABC, DEF are equal in all respects (4. 1.). Wherefore, if two triangles &c. Q. E. D.

PROPOSITION IX. PROBLEM.

To bisect a given rectilineal angle; that is, to divide it into two equal angles.

Let BAC be the given rectilineal angle; it is required to

bisect it.

Take any point D in AB, and from AC cut off AE equal to AD (3. 1.); join DE, and on it describe an equilateral triangle DEF (1. 1.); join AF; then the straight line AF bisects the angle BAC.

Because AD is equal to AE, and AF is common to the two triangles DAF, EAF, and DF is equal to EF; the angle DAF is equal to EAF (8. 1.); wherefore the angle BAC is bisected by the straight line AF. Which was to be done.

A

Scholium. In the same manner may each of the angles BAF, CAF be bisected. Therefore by successive subdivisions an angle may be divided into four, eight, sixteen, &c. equal parts.

ED.

PROPOSITION X. PROBLEM.

To bisect a given finite straight line; that is, to divide it into two equal parts.

Let AB be the given straight line; it is required to divide it into two equal parts.

On AB describe an equilateral triangle ABC (1. 1.), and bisect the angle ACB by the straight line CD (9. 1.); AB is bisected in the point D.

C

Because AC is equal to CB, and CD is common to the two triangles ACD, BCD, and the angle ACD is equal to BCD; the base AD is equal to DB (4. 1.); therefore the straight line AB is divided into two equal parts in the point D.

D

B

Cor. If a straight line bisect the vertical angle of an equilateral or an isosceles triangle, it will also bisect the base at right angles. For the triangles ACD, BCD are equal in all respects.

PROPOSITION XI. PROBLEM.

To draw a straight line at right angles to a given straight line, from a given point in the same.

Let AB be a given straight line, and C a point given in AB; it is required to draw a straight line from C at right angles to AB.

F

Take any point D in AC, and make CE equal to CD (3. 1.); on DE describe the equilateral triangle DFE (1. 1.), and join FC; the straight line FC, drawn from the given point C, is at right angles to the given straight line AB.

Because DC is equal to CE, and FC is common to the two triangles DCF, ECF, and the A side DF is equal to EF; the an

E

gle DCF is equal to ECF (8. 1.); and they are adjacent angles; therefore each of the angles DCF, ECF, is a right angle (10 Def.). Wherefore from the given point C, in the given straight line AB, FC has been drawn at right angles to AB.

Note. If the given point C be at or near the end of the given line AB, then AB must be produced on the side of the point C.

ED.

PROPOSITION XII. PROBLEM.

To draw a straight line perpendicular to a given straight line of unlimited length, from a given point without the line.

Let AB be a given straight line, which may be produced to any length both ways, and let C be a given point without the line; it is required to draw from C a straight line perpendicular to AB.

Take any point Don the other side of AB, and from the center C, with the radius CD, describe the circle EGF meeting AB in F and G (3 Post.); bisect FG in H (10. 1.), and draw CF, CH, CG; the A straight line CH, drawn from the given point C, is

perpendicular to the given straight line AB.

Because FH is equal to HG, and HC is common to the two triangles FHC, GHC, and the side CF is equal to CG (20 Def); the angle CHF is equal to CHG (8. 1.); and they are adjacent angles; therefore each of them is a right angle, and CH is perpendicular to FG (10 Def.). Therefore from the given point C a perpendicular CH has been drawn to the given straight line AB.

PROPOSITION XIII. THEOREM.

The two angles which one

straight line makes

with another, on one side of it, are together equal to two right angles.

Let the straight line AB make with CD, on one side of CD, the angles ABC, ABD; these are together equal to two right angles.

If the angle ABC be equal to ABD, each of them is a right angle (Def. 10.). But, if not, from the point B draw BE at right angles to CD (11. 1.); then the angles CBE, EBD are two right angles. The angle DBA is equal to the two angles DBE, EBA. To these equals add the angle CBA, then the two angles DBA, CBA will be equal to the three angles DBE, EBA, ABC. Now DBE is a right angle, and the two angels EBA, ABC are together equal to the right angle CBE. Therefore the sum of the three angles DBE, EBA, ABC is equal to two right angles. Consequently the sum of the two angles DBA, CBA is also equal to two right angles. Wherefore, the angles &c. Q. E. D. LEGENDRE.

Otherwise. The construction remaining, the angles CBE, EBD are right angles. But the angle ABD is greater than EBD by the angle ABE, and the angle ABC is less than EBC by the same angle ABE. Therefore the angles ABD, ABC are together equal to the angles CBE, EBD, that is, to two right angles. Wherefore, the angles &c. Q. E. D. ED.

Cor. 1. All the angles which any number of straight lines make with another straight line at the same point, and on the same side of the line, are together equal to two right angles. For their sum is equal to the sum of the two angles BA, CBA.

ED.

Cor. 2. If two straight lines cut each other, the four angles which they make at the point of intersection are together equal to four right angles.

For the two angles on each side of AB are together equal to two right angles, therefore the four angles on both sides of AB are together equal to four right angles.

ED.

Cor. 3. All the angles made by any number of straight lines, which intersect or meet one another in one point, are together equal to four right angles.

F