double of the angle C. Show that the side CB is double of the side AB.

B

Produce BA to D, making AD equal to AB (1. 3), and join CD.

Proof. In the triangles DAC, BAC, the side DA is equal to the side BA (con.),

and the side AC is common to the two triangles;

therefore the two sides DA, AC are equal to the two sides BA, AC, each to each,

and the included angle DAC is equal to the included angle BAC, being right angles (1. 13, and ax. 11);

therefore the base DC is equal to the base BC (1. 4),

and the triangle DAC is equal to the triangle BAC in every respect;

therefore the angle ADC is equal to the angle ABC,

and the angle ACD is equal to the angle ACB:

wherefore the angle BCD is double the angle BCA, and therefore equal to the angle ABC (ax. 6). Therefore the triangle BDC is equiangular; wherefore it is equilateral (1. 6, Cor.).

Therefore the side CB is equal to the side BD: but the side BD is double of BA (con.);

wherefore the side CB is double of the side BA. Q.E.D.

33. ABCD is a parallelogram whose diagonals AC, BD intersect in O: show that if the parallelograms AOBP,

DOCQ be completed, the straight line joining P and Q passes through O.

P

A

D

B

Join PO, QO.

Then POQ shall be a straight line.

Proof-The diagonals of a parallelogram bisect each other; for the angles OAD, ADO are equal to the angles OCB, CBO (I. 29),

and the side AD is equal to the side CB (1. 34),

therefore AO is equal to CO, and DO is equal to BO (1. 26). Because DO is equal to OB,

and OB is equal to AP (1. 34),

therefore DO, AP are both equal and parallel;

wherefore PO, AD are both equal and parallel (1. 33). Similarly it may be proved that QO, AD are both equal and parallel.

In the triangles PAO, ODQ,

the side AO is equal to the side DQ (1. 34),

and the side OP is equal to the side QO (ax. 1), both being equal to AD;

therefore the two sides AO, OP are equal to the two sides DQ, QO, each to each,

and the base AP is equal to the base DO;

therefore the angle AOP is equal to the angle DQO (1. 8): to each of these equals add the angle AOQ;

therefore the angles AOP, AOQ are equal to the angles DQO, QOA (ax. 2):

but the angles DQO, QOA are together equal to two right angles (1. 29);

therefore the angles AOP, AOQ are together equal to two right angles (ax. 1):

wherefore POQ is a straight line (1. 14). Q.E.D.

34. Prove that any line terminated by two opposite sides of a parallelogram, and passing through the intersection of the diagonals, is bisected in that point.

Let ABCD be a parallelogram, of which AC, BD are the diagonals.

Then every straight line drawn through O, the intersection of AB, CD, and terminated by two opposite sides of the parallelogram, shall be bisected in O.

Proof-Through O draw the straight line EF, terminated by AD, BC.

Because AD is parallel to BC, and AC meets them,

therefore the angle EAO is equal to the alternate angle FCO (1. 29),

and the angle AOE is equal to the angle COF (1. 15).

Hence in the triangles AOE, COF,

the two angles EAO, AOE are equal to the two angles FCO, COF, each to each,

and the side AO is equal to the side CO, the diagonals of a parallelogram bisecting each other (see deduction to L. 33); therefore the side EO is equal to the side FO (1. 26): wherefore EF is bisected in O.

Therefore every straight line drawn through O, and terminated by the sides of the parallelogram, that is, by AD, BC, or AB, DC, is bisected in O. Q.E.D.

The line drawn from the right angle of a triangle to the middle of the hypotenuse is equal to half the hypotenuse.

Let ABC be a right-angled triangle, having the angle at A

a right angle, and let AD be drawn to the point of bisection of BC.

Then AD shall be equal to half BC.

B

Produce AD to E, making DE equal to AD (1. 3), and· join BE.

Proof-In the triangles ADC, BDE,

the side AD is equal to the side DE (con.),

and the side CD is equal to the side BD (hyp.);

therefore the two sides AD, DC are equal to the two sides

ED, DB, each to each;

and the angle ADC is equal to the angle EDB (1. 15); therefore the base AC is equal to the base EB (1. 4), and the triangles are equal in every respect;

therefore the angle CAD is equal to the angle BED.

Because AE meets the two straight lines AC, BE, and makes the alternate angles CAE, AEB equal to one another,

therefore AC is parallel to BE (1. 27).

Because AC is parallel to BE, and AB meets them,

the two interior angles CAB, ABE are together equal to two right angles (1. 29);

but the angle CAB is a right angle (hyp.); therefore the angle ABE is a right angle.

Hence in the triangles BAC, ABE,

the two sides BA, AC are equal to the two sides AB, BE, each

to each,

and the angle BAC is equal to the angle ABE (ax. 11); therefore the base BC is equal to the base AE (1. 4);

C

and AD is half of AE (con.);

wherefore AD is equal to half BC. Q.E.D.

If the two diagonals are drawn, show that a parallelogram will be divided into four equal parts; in what case will the diagonal bisect the angle?

Let ABCD be a parallelogram, of which AC, BD are the diagonals.

Then the parallelogram ABCD shall be divided into four equal parts.

B

Let O be the point of intersection of the diagonals.

Proof. The diagonals of a parallelogram bisect each other. (See deduction to I. 33.)

Because AO is equal to OC,

the triangle AOD is equal to the triangle COD (1. 38); but the triangle ACD is half the parallelogram ABCD (1. 34); therefore each of the triangles AOD, COD is a quarter of the parallelogram ABCD.

Similarly it can be proved that each of the triangles AOB, COB is a quarter of the parallelogram.

Wherefore the parallelogram ABCD is divided into four equal parts by the diagonals AC, BD. Q.E.D.

Note. The diagonals bisect the angles when the parallelogram is equilateral.

Let ABCD be an equilateral parallelogram.