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Proposition 47 is said to have been discovered by Pythagoras, a Greek, who lived about 550 B.C. It is sometimes called The Theorem of Pythagoras.

Many proofs have been invented of this important proposition.

The proof given on pages 148, 149 is said to be due to Euclides, a Greek of Alexandria, who about the year 300 B.C. wrote the Treatise on Geometry etc. which bears his name We give below figures which will suggest two methods of proving Prop. 47.

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The above figure shews how from the two squares to cut off two triangular areas which, being placed differently, change the two squares into one.

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The above two figures shew how by taking away four equal triangular areas from a certain square we may either have two squares or a single square left.




Proposition 31.

147. To draw a straight line through a given point parallel to a given straight line.

Let A be the given point, BC the given straight line ; it is required

to draw through A a straight line parallel to BC.




Construction. In BC take a point D.

Join AD.

At the point A in the straight line AD,

make the angle DAE equal to the angle ADB. [Prop. 23] Then AE is the line required.


The line AD makes with the lines EA, DC the

alternate angles EAD, ADB ;

and because these alternate angles are equal, therefore the line AE is parallel to DC, [Prop. 27]

and it is drawn through the point A. Wherefore, a straight line has been drawn, etc.


Example. Divide a given straight line into any given number (say five) of equal parts.

Let AB be the given straight line;

it is required to divide AB into five equal parts.

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Construction. From A draw a line AC making an angle of about half a right angle with AB.

Take a point D in AC such that AD is about a fifth part of AB; make DE, EF, FG, GH each equal to AD.

Join BH.

Through G, F, E, D draw GS, FR, EQ, DP each parallel to BH, cutting AB in S, R, Q, P respectively.


Proof. Because AQE is a triangle and AE is bisected and DP drawn parallel to QE, therefore AP=PQ. [Example ii, p. 123.] Similarly by drawing through D a line parallel to PR,

we may prove that PQ=QR.

Similarly we may prove that QR=RS=SB.

Wherefore, the given line AB has been divided into the five equal parts AP, PQ, QR, RS, SB. Q.E.F.


1. Describe a parallelogram having two adjacent sides equal to two given finite straight lines and one of its angles equal to a given angle.

2. Construct a rectangle having its sides equal respectively to two given finite straight lines.

3. Through three given points draw three straight lines forming a triangle equiangular with a given triangle.

4. Divide a given parallelogram into a given number of parallelograms of equal areas, each equiangular with the given parallelogram.

5. On a given straight line describe a rhombus having an angle equal to a given angle.

6. Draw through a given point a straight line to make equal angles with two given finite straight lines which are not parallel, but which do not meet within the limits of your diagram.

Proposition 42.

148. To describe a parallelogram equal in area to a given triangle, having one of its angles equal to a given angle.

Let ABC be a given triangle, D a given angle ;

it is required to describe a parallelogram whose area is equal to that of ABC, such that one of its angles is equal to the given angle D.


Construction. Bisect BC in E.

[Prop. 10]

At the point E in EC make the angle CEF equal to the

angle D.

[Prop. 23]

[Prop. 31]

[Prop. 31]

Through C draw CG parallel to EF. Through A draw AFG parallel to BC. Then, FECG is the parallelogram required. Proof. Since the opposite sides of CEFG are parallel, therefore CEFG is by construction a parallelogram; because the parallelogram FECG is on the same base EC and between the same parallels, as the triangle AEC, therefore the area FECG is double of AEC. [Prop. 41] Again, because BE is equal to EC,

therefore the areas ABE, AEC are equal; [Prop. 37] so that the area ABC is double of AEC;

the doubles of equals are equal;

so that the area FECG is equal to the area ABC,
and FECG is a parallelogram,

one of whose angles FEC is equal to D.
Wherefore, a parallelogram equal in area, etc.


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