Proof-The sides AB, BC are greater than the side AC

(1. 20);

the sides AD, DC are greater than the side AC;

the sides BA, AD are greater than the side BD; and the sides BC, CD are greater than the side BD; therefore twice the sides AB, BC, CD, DA are together greater than twice the diagonals AC, BD;

wherefore the sides AB, BC, CD, DA are together greater than the diagonals AC, BD. Q.E.D.

3. Through a given point within an angle BAC to draw a straight line cutting off equal parts from AB, AC.

Let D be the given point within the angle BAC.

It is required to draw through D a straight line which shall cut off equal parts from AB, AC.

H

Produce BA to E (post. 2), and bisect the angle CAE by the straight line AF (1. 9);

through D draw GH parallel to AF, cutting AB, AC in G, H respectively (1. 31).

Then AG shall be equal to AH.

Proof.-Because GH is parallel to AF, and EB falls upon them,

the exterior angle EAF is equal to the interior and opposite angle AGH (1. 29);

and because GH is parallel to AF, and AH meets them,

the angle FAH is equal to the alternate angle AHG (L. 29):

but the angle EAF is equal to the angle FAH (con.), therefore the angle AGH is equal to the angle AHG; wherefore AG is equal to AH (1. 6). Q.E.D.

4. By the method of the first proposition describe on a given finite straight line an isosceles triangle the sides of which shall be each equal to twice the base.

Let AB be the given straight line.

It is required to describe an isosceles triangle upon AB, the equal sides of which shall be twice AB.

E H

From the centre A, at the distance AB, describe the circle BCD (post. 3);

and from the centre B, at the distance BA, describe the circle ACE;

produce the base AB both ways to meet the circles in D, E (post. 2):

from the centre A, at the distance AE, describe the circle

EFG;

and from the centre B, at the distance BD, describe the circle

DFH;

from F, one of the points in which the larger circles cut one another, draw the straight lines FA, FB to the points A, B (post. 1).

Then ABG shall be the isosceles triangle required.

Proof. Because A is the centre of the circle BCD, therefore AD is equal to AB (def. 15);

wherefore BD is double of AB:

similarly it can be proved that AE is double of AB. Because A is the centre of the circle EFG,

therefore AF is equal to AE;

but AE is double of AB;

therefore AF is double of AB:

similarly it can be proved that BF is double of AB: and because AF, BF are both double of AB,

therefore AF is equal to BF (ax. 6).

Wherefore upon the given base AB an isosceles triangle ABG has been described, the equal sides of which, viz. AF, BF, are twice AB.

Q.E.F

5. Describe on a given straight line, as diagonal, a rhombus, each of whose sides is equal to that line, by the method employed in the first proposition.

Let AB be the given straight line.

It is required to describe on AB, as diagonal, a rhombus, each of whose sides shall be equal to AB.

From the centre A, at the distance AB, describe the circle BCD (post. 3);

from the centre B, at the distance BA, describe the circle ACD;

and from the points C, D, where the circles cut one another, draw the straight lines CA, CB, DA, DB (post. 1).

Then ACBD shall be the rhombus required.

Proof. Because A is the centre of the circle ACD, therefore AC, AD are each equal to AB (def. 15);

because B is the centre of the circle BCD,

therefore BC, BD are each equal to BA;

and because AC, AD, BC, BD are each equal to AB, therefore they are equal to one another (ax. 1).

Wherefore ACBD is a rhombus (def. 30), and each side is equal to AB. Q.E.F

6. If two right-angled triangles have two sides containing an acute angle of the one equal to two sides containing an acute angle of the other, show that the triangles are equal in all respects.

Let ABC, DEF be two right-angled triangles having the angles at B, E right angles,

and the sides BA, AC equal to the sides ED, DF, each to each.

Then the triangles ABC, DEF shall be equal in all respects.

Proof. For if BC be not equal to EF,
one of them must be greater than the other.
If possible, let BC be greater than EF,
make BG equal to EF (1. 3), and join AG.

Then in the triangle ABG, DEF,

the two sides AB, BG are equal to the two sides DE, EF,

each to each,

and the angle ABG is equal to the angle DEF (ax. 11),

therefore the base AG is equal to the base DF (1. 4);

but AC is equal to DF (hyp.);

therefore AG is equal to AC (ax. 1).

Because AG is equal to AC,

the angle AGC is equal to the angle ACG (1. 5);

but the angle AGC is greater than the angle ABG (1. 16), that is, greater than a right angle,

wherefore the angle ACG is also greater than a right angle; therefore the angles AGC, ACG are together greater than two right angles, which is impossible (1. 17);

wherefore BC is not unequal to EF, that is, BC is equal to EF, and also the remaining angles of the one triangle to the remaining angles of the other (1. 8). Q.E.D.

7. If a point P be taken inside a quadrilateral ABCD, prove that the sum of the distances of P from the angular points is the least possible when P is situated at the intersection of the diagonals.

Let P be the intersection of the diagonals AC, BD. Take any other point O inside the quadrilateral ABCD. Then the sum of the distances PA, PB, PC, PD shall be less than the sum of the distances OA, OB, OC, OD.

Proof. The sides OA, OC are greater than the side AC (1. 20),

and the sides OB, OD are greater than the side BD; therefore OA, OB, OC, OD are together greater than AC, BD, that is, greater than PA, PB, PC, PD. Q.E.D.

8. The sum of the diagonals of a quadrilateral is less than the sum of any four lines that can be drawn from any point whatever (except the intersection of the diagonals) to the four angles.

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