but the angle HBG is equal to the angle GBC (con.); therefore the angle HGB is equal to the angle HBG (ax. 1); wherefore HG is equal to HB (1. 6);

but DE is equal to HG (1. 34);

therefore DE is equal to HB.

Wherefore DE is equal to the difference between BD, CE. Q.E.F.

32. If the straight line bisecting the exterior angle be parallel to a side, show that the triangle is isosceles.

Let the straight line CE bisecting the exterior angle ACD of the triangle ABC be parallel to the side AB.

Then the triangle ABC shall be isosceles.

Proof. Because CE is parallel to AB, and AC meets them, the angle ACE is equal to the alternate angle BAC (1. 29); and because CE is parallel to AB, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC (1. 29);

but the angle ACE is equal to the angle ECD (hyp.); therefore the angle BAC is equal to the angle ABC; and therefore the side CA is equal to the side CB (1. 6). Wherefore the triangle ABC is isosceles. Q.E.D.

If the interior angle at one angular point of a triangle and the exterior angle at another be bisected by straight lines, the angle contained by the two bisecting lines is equal to half the third angle of the triangle.

Let ABC be a triangle, and let the interior angle ABC be bisected by the straight line BE, and the exterior angle ACD by the straight line CE.

Then the angle BEC shall be equal to half the angle BAC.

Proof. The exterior angle ACD is equal to the two interior and opposite angles CBA, BAC (1. 32);

the exterior angle ECD is equal to the two interior and opposite angles CBE, BEC;

but the angle ECD is half the angle ACD (hyp.);

therefore the angles CBE, BEC are equal to half the angles СВА, ВАС;

but the angle CBE is half the angle CBA (hyp.);

wherefore the angle BEC is equal to half the angle BAC. Q.E.D.

Show by the proposition that if two straight lines be perpendicular to two other straight lines, each to each, the angle included by the first pair is equal to the angle included by the second.

Let AB, CD be two straight lines intersecting in E, and let FG, HK, intersecting in L, be perpendicular to AB, CD respectively.

Then the angle CEB shall be equal to the angle FLK.

Proof-The angles of a quadrilateral are together equal to four right angles,

because by drawing a diagonal it is divided into two triangles (I. 32);

therefore the angles HEF, EFL, FLH, LHE are together equal to four right angles :

but the angles EFL, LHE are right angles (hyp.);

therefore the angles HEF, FLH are together equal to two right angles (ax. 3):

the angles FLK, FLH are together equal to two right angles (1. 13);

therefore the angles HEF, FLH are equal to the angles FLK, FLH (ax. 1);

wherefore the angle HEF, that is, the angle CEB, is equal to the angle FLK (ax. 3). Q.E.D.

If the base of a triangle be produced both ways, the exterior angles are greater than two right angles by the vertical angle.

Let ABC be a triangle, and let the base BC be produced both ways to D and E. Then the exterior angles ABD, ACE shall be greater than two right angles by the vertical angle BAC.

Proof. The exterior angle ABD is equal to the two interior and opposite angles ACB, BAC (1. 32);

and the exterior angle ACE is equal to the two interior and opposite angles ABC, BAC;

therefore the two exterior angles ABD, ACE are together equal to the angles ABC, ACB, and twice the angle BAC; but the three angles ABC, ACB, BAC are together equal to two right angles (1. 32);

wherefore the exterior angles ABD, ACE are greater than two right angles by the vertical angle BAC. Q.E.D.

If the three sides of a triangle be produced, the sum of the exterior angles is equal to four right angles.

Let ABC be a triangle, and let the sides AB, BC, CA be produced to D, E, F respectively.

Then the exterior angles CBD, ACE, BAF shall be together equal to four right angles.

Proof. The exterior angle CBD is equal to the two interior and opposite angles ACB, BAC (1. 32);

the exterior angle ACE is equal to the two interior and opposite angles ABC, BAC;

and the exterior angle BAF is equal to the two interior and opposite angles ACB, ABC;

therefore the three exterior angles CBD, ACE, BAF are together equal to twice the angles ABC, BCA, CAB: but the angles ABC, BCA, CAB are together equal to two right angles (1. 32);

wherefore the three exterior angles CBD, ACE, BAF are together equal to four right angles. Q.E.D.

If the alternate sides of any polygon be produced to meet, the angles formed by these lines, together with eight right

angles, are together equal to twice as many right angles as the figure has sides.

Proof. The exterior angle ACD is equal to the two interior and opposite angles CBA, BAC (1. 32);

but the interior angle BCE is equal to the angle ACD (1. 15), the angle CBA is an exterior angle of the polygon,

and the angle BAC is an angle formed by two alternate sides of the polygon produced;

therefore all the interior angles of the polygon are equal to all the exterior angles of the polygon together with the angles formed by the alternate sides produced:

but all the exterior angles of the polygon are equal to four right angles (1. 32, Cor. 2);

therefore the angles formed by the alternate sides produced, together with four right angles, are equal to all the interior angles of the polygon :

but all the interior angles, together with four right angles, are equal to twice as many right angles as the figure has sides (1. 32, Cor. 1);

wherefore the angles formed by the alternate sides produced, together with eight right angles, are together equal to twice as many right angles as the figure has sides. Q.E.D.

ABC is a triangle right-angled at A, and the angle B is