PROBLEM (1).

To describe a triangle having two of its sides equal to two given straight lines, and the opposite to one of them equal to a given L.

Let A, B be the given straight lines and C the given 4. It is required to describe a ▲ having two of its sides equal to A, B and the opposite the latter = C.

Make SRT: ▲ C, and cut off RS = A,

and with centre S and radius B describe a O.

Then the

=

If this meet RT in T, join ST.

RST will be one which satisfies the conditions of the question.

N.B. When C is either an obtuse or a rt. < the problem will admit of one solution if B be > A ; otherwise the problem will be impossible.

When C is an acute angle and B= the perpendicular from S on RT, then only one straight line can be drawn from SB, and there will thus be one solution.

If B be < the perpendicular from S on RT, the problem is impossible.

If B be the perpendicular from S on RT, then two straight lines can be drawn from S to RT, each = B, which will be on the same side of SR if B be < A, and thus give two solutions; otherwise, if B be not < A, only one solution.

THEOREM (m).

If a straight line be divided into two equal parts and also two unequal parts, the squares on the unequal parts shall together be twice the squares on half the line and on the line between the points of section.

Let AH be divided into two equal parts in C, and into two unequal parts in D;

then shall the squares on AD, DH be = twice the squares on AC, CD.

=

We shall first describe a square twice the squares on AC, CD.

Draw CF and = AC; join AF.

Then ACF is a right-angled isosceles ▲, and square on AF twice the square on AC.

=

(I. 35) Join FH; draw DK || to CF; KG || to AH; join AK. Then it may be easily shewn that FCH, KDH, FGK are right-angled isosceles ▲ s, and that ▲ AFH is a right ‹ . Hence square on FK twice the square on GK, (1. 35) twice the square on CD; (1. 26) twice the squares on AC, CD; twice the squares on AC, CD. squares on AD, DK,

.. squares on AF, FK ... square on AK

Also squares on AD, DH

and ...

=

=

=

=

=

=

square on AK;

(I. 35)

twice the squares on AC, CD.

COR. Let AH be bisected in C and produced to D: then shall the squares on AD, DH= twice the squares on AC, CD. This may be established in a similar manner to the above or deduced from it, thus:

Produce DA and make AX = HD. Then XC CD.

=

=

... squares on XH, HD twice the squares on XC, CH, i.e. squares on AD, HD = twice the squares on DC, CA.

THEOREM (2).

The squares on the sides of a triangle are double the squares on half the base and on the line joining the vertex and the middle point of the base.

From the vertex K of ▲ AKH draw KD 1 to AH.

Bisect AH in C, and join CK.

Then shall the squares on AK, KH = twice the squares on AC, CK.

For the squares on AD, DH = twice the squares on AC, CD;

.. the squares on AD, DH and twice the square on DK= twice the square on AC and twice the squares on CD, DK;

i.e. the squares on AK, KH = twice the squares on AC, CK.

Similarly, if the perpendicular from K fall on AH pro

duced.

K

A

C

Since the squares on AD, DH = twice the squares on AC, CD, it follows that the squares on AK, KH = twice the squares on AC, CK.

Euclid solves the following two problems, the necessity for which has been removed in this treatise by postulate 3.

The corresponding postulate of Euclid is "Let it be granted that a circle may be described from any centre at any distance from that centre," which does not allow the transfer of distances by means of compasses.

PROBLEM).

From a given point to draw a straight line equal to a given straight line.

With centre Q and radius QR describe a C cutting QC produced in X.

Then CX shall be the line required.

For QX
QA is

=

Q is centre of circle RX;

QC; . QAC is an equilateral ;

... ▲

.. CX AR; but AR = AB;

=

.. CX= AB;

.. from C a straight line CX has been drawn = AB.

Q. E. F.

PROBLEM (9).

From the greater of two straight lines to cut off a part

equal to the less.

B

X

D

Let AB CD, be the two given straight lines of which CD is the greater.

It is required to cut off from CD a part

From C draw a straight line CX = AB.

With centre C and radius CX describe a circle cutting

CD in Z.

Then CZ is the part required.

For CZ CX, and CX = AB ;.

.:. CZ = AB.

... from CD a part CZ has been cut off = AB.

Q. E. F.