DEFINITIONS.

An obtuse angle is one greater than a right angle. An acute angle is one less than a right angle. An obtuse-angled triangle is one having an obtuse angle.

An acute-angled triangle is one having three acute angles.

PROPOSITION III.

In obtuse-angled triangles, the square on the side opposite the obtuse angle is greater than the squares on the sides containing it by twice the rectangle contained by either of those sides and the part of it intercepted between the perpendicular let fall upon it from the opposite angle and the obtuse angle.

B

Let ABC be an obtuse-angled ▲, having the obtuse LACB,

and from B let fall BK 1 to AC produced.

Then shall the square on AB be > the squares on AC, CB by twice the rectangle AC, CK.

For AK is the sum of AC and CK; ... square on AK is = squares on AC, CK

... square on AB is = squares on AC, CB

+ twice rectangle AC, CE.

PROPOSITION IV.

In any triangle the square on the side opposite to any one of the acute angles is less than the squares on the sides containing it by twice the rectangle contained by either of those sides and the part of it intercepted between the perpendicular let fall upon it from the opposite angle and the acute angle.

AAA

Let ABC be a ▲ having the acute ▲ ACB, and from B let fall BK 1 CA or CA produced. Then shall the square on AB be < the squares on AC, CB by twice the rectangle AC, CK.

For should the

fall within or without the ▲ ABC, in either case AK is the difference between AC and CK.

=

.. square on AK + twice rectangle AC, CK is

on AC, CK.

= squares

(II. 2)

.. squares on AK, KB + twice rectangle AC, CK= squares on AC, CK, KB.

.. square on AB + twice rectangle AC, CK = squares on AC, CB.

Should the BAC be a right ≤, and ... BK coincide with BA,

then square on AB + twice square on AC = squares on AC, CB. (1.35)

PROBLEM A.

Describe a square equal to a given rectilineal figure.

Let A be the given rectilineal figure.
First describe the rectangle FH = A;
then if the sides are equal FH is a square,
and what was required is done.

But, if not, produce FG and cut off GM = GH.
Bisect FM in P,

and with centre P and radius PM describe o FQM.
Produce HG to meet the O in Q.

=

Join PQ.

Then. FG is the sum and GM = the difference of PM and PG;

... rectangle FG, GM is = the difference of the squares on PM and PG,

(II. 2)

the difference of the squares on PQ and PG, the square on GQ,

rectangle FG, GM, for GH = GM,

and FH was made = A ;

.. square on GQ = A.

(1.35)

PROBLEM B.

To divide a straight line into two parts so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.

Let PQ be the given straight line. On PQ describe the square PR. Bisect PS in K and join QK. Produce SP and cut off KG = KQ.

On PG describe a square GPCF, one side of which PC will fall on PQ. Then shall PQ be divided in C, so that rectangle PQ, QC= square on PC.

K

G

F

H

R

... rectangle SG, GP is the difference of the squares

on KG and KP,

and is ... =

(II. 2)

the difference of the squares on KQ and KP,

and is ... =

the square on PQ;

(1.35)

but figure GH = rectangle SG, GP; for GF= GP,
and figure PR is the square on PQ;

.. figure GH = figure PR.

Take away the common part, the figure PH,

... the remainder GC the remainder CR,

=

but GC is the square on PC and CR is = rectangle PQ, QC, for QR is = PQ;

... rectangle PQ, QC = square on PC.

PROBLEM C.

To describe an isosceles triangle having each of the angles at the base double of the angle at the vertex.

Take any straight line AB and

From let fall QP1 to BC, and.. bisecting BC.

Then squares on AB, BQ are together > square on AQ by

twice rectangle AB, BP.

(II. 4)

But rect. AB, BC = rect. AB, BP + reċt. AB, PC and ..twice the rect. AB, BP;

... square on AC = twice rectangle AB, BP;

(II. I)

and is .. = LS CQA, CAQ;

and is ... double of ▲ CAQ.

Thus each of the angles at the base BQ of A ABQ is

double of the angle at the vertex A.