PROPOSITION XVI. THEOR.

If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles CBA or BAC.

A

E

F

G

Bisect (15) AC in the point E; join BE and produce it to F, and make EF equal to EB; join FC, and produce AC to G. Because the side AE is equal EC and BE to EF, and the included angles AEB and CEF are equal, because (7) they are verti- B cal angles; the triangles ABE and CFE are (12) equal; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater (Ax. 8) than ECF; therefore the angle ECD or ACD is greater than the angle BAE or BAC. In the same manner, if the side BC be bisected, it might be demonstrated that the angle BCG, that is (7) the angle ACD, is greater than the angle ABC.

PROPOSITION XVII. THEOR.

Any two angles of a triangle are together less than two right angles.

Let ABC be any triangle, any two of its angles are together less than two right angles.

Produce BC to D, and because ACD is the exterior angle of the triangle ABC, ACD is greater (16) than the interior opposite angle ABC; to each of these add the angle ACB; therefore the angles ACD, ACB, are greater B than the angles ABC, ACB; but ACD and

A

ACB are together equal (5) to two right angles; therefore

the angles ABC, ACB, are together less than two right angles. In like manner, it may be demonstrated that the angles BAC, ACB, as also CAB, ABC, are together less than two right angles.

Cor. A triangle can have only one right or one obtuse angle, and two of its angles are always acute.

PROPOSITION XVIII. THEOR.

The greater side of every triangle has the greater angle opposite to it.

Let ABC be a triangle, of which the side AC is greater than AB, the angle ABC is also greater than the angle BCA. From AC, which is greater than AB, cut off

B

(2) AD, equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, it is greater (16) than the interior opposite angle BCD; but ADB is equal ABD, because the side AB is equal to AD; therefore the angle ABD is also greater than the angle ACB; wherefore much more is the angle ABC greater than ACB.

Cor. Hence the less angle of every triangle is opposite to the less side.

PROPOSITION XIX. THEOR.

If two angles of a triangle are equal, the sides opposite to them are also equal.

Let ABC be a triangle having the angle B equal to the angle C, the side AB is also equal to the side AC.

For if it be not equal, AB must either be greater than AC or less than it; but AB is not greater than AC, for then the angle C would be greater (18) than B, which it is not; therefore AB is not greater than AC, neither is it less, for then the angle C would be less (18 Cor.) than B,

B

A

C

which it is not; therefore AB is not less than AC; and it has been proved that it is not greater; therefore AB is equal to AC.

The greater angle of every triangle has the greater side opposite to it.

A

Let ABC be a triangle of which the angle B is greater than the angle C, the side AC is also greater than the side AB. For if it be not greater, AC must either be equal to AB or less than it; it is not equal, because then the angle B would be equal (14) to the angle C; but it is not, therefore AC is not equal to AB; neither is it less, for then the angle B would be less than C, but it is not; therefore the side AC is not less than AB; and it has been proved that it is not equal to AB; therefore AC is greater than AB.

B

If from the ends of one side of a triangle there be drawn two straight lines to a point within the triangle, these two straight lines are less than the two sides of the triangle, but contain a greater angle.

Let the two straight lines BD and CD be drawn from B and C, the ends of the side BC of the triangle ABC, to the point D within it, BD and DC are less than the other two sides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC.

Produce BD to E; and because two sides of a triangle (8) are greater than the third side, the two sides BA, AE of the triangle ABE are greater than BE; to each of these add EC, then the sides BA, AC are greater than BE, EC. B

A

E

Again, because the two sides CE, ED of the triangle CED are

greater than CD, if BD be added to each, the sides CE, EB are greater than CD, DB; but it has been shown that BA, AC are greater than BE, EC, much more then are BA, AC greater than BD, DC. Again, because the exterior angle of a triangle is (16) greater than the interior opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDC is greater than the angle CEB, much more then is the angle BDC greater than the angle BAC.

PROPOSITION XXII. PROB.

From a given point in a straight line to draw a perpendicular to it.

From C, the given point in the straight line AB, it is required to draw a perpendicular to AB.

F

B

Take any point D in AB, and (2) make CE equal CD, and on DE describe (10) the isosceles triangle DFE, and join FC, the straight line FC is perpendicular to AB. A Because in the triangles DFC and EFC the D C E side DC is equal to CE, FD to FE, and the side FC common to both, the triangles are (11) equal; and therefore the angle DCF is equal to ECF, and they are adjacent angles; wherefore (Def. 9) each of them is a right angle, and the straight line FC is perpendicular to AB.

PROPOSITION XXIII. PROB.

From a given point without a given straight line to draw a perpendicular to it.

Let AB be the given straight line, and C a point without it. It is required to draw from the point C a straight line perpendicular to AB.

A F

D

G

Take any point D on the other side of AB, and from C as a centre, with the radiant CD, describe (Post. 3) the circle EGF, meeting AB in F and G, bisect (15) FG in H, and join CF, CH, CG, the straight line CH drawn from the point C is perpendicular to AB. Because in the triangles FCH and GCH the side FH is equal to HG, CF to CG, and CH is common to both; therefore the triangles (11) are equal; the angle CHF is equal to CHG, and they are adjacent angles; wherefore (Def. 9) each of them is a right angle, and CH is perpendicular to AB.

PROPOSITION XXIV. THEOR.

If from the middle point of a straight line a perpendicular to it be drawn, every point of the perpendicular is equally distant from the extremities of the line, and every point that is not in the perpendicular is unequally distant from the same extremities.

Let C be the middle point of the straight line AB, and let CD be drawn perpendicular to it, then any point E in DC is equally distant from A and B.

Join AE and EB. In the triangles ACE and BCE the side AC is equal to CB, and CE is common to both, and the included angle ACE is equal to the included angle BCE; therefore (12) AE is equal to BE. Again,

any point F that is not in the perpendicular is A C B unequally distant from A and B. Draw AF, cutting CD in G, and join FB and GB. Because, from what has been already proved, AG is equal to BG; and adding GF to each, AF is equal to BG, GF; but BG, GF are together (8) greater than BF; hence AF is greater than BF, and the point F is unequally distant from A and B.