so that, the angle BAC is equal to
the corresponding angle DAC;

and DAC is a right angle;

so that BAC is a right angle.

Wherefore, when the area of the square on one side, etc.

Q.E.D.

EXAMPLES L.

1. ABC, DBC are two right-angled triangles having a common hypotenuse BC; prove that if AB is greater than DB then AC is less than DC.

2. If the difference of the areas of the squares on two sides of a triangle is equal to the area of the square on the third side the triangle is right-angled.

3. Part of the locus of a point P which is such that the difference of (the areas of) the squares described on the distances of P from two given points A, B is constant, is a straight line, perpendicular to AB. [See Question 9, Examples XLIX.]

4. Two isosceles triangles have their equal sides of the same length; prove that if they are of the same altitude they are equal in all respects.

5. Two isosceles triangles have their equal sides of the same length; prove that the triangle which has the greater altitude has the shorter base.

6. If the sum of the (areas of the) squares on two opposite sides of a quadrilateral is equal to the sum of (the areas of) the squares on the other two sides, the diagonals of the quadrilateral are at right angles. [See Question 12, Examples XLIX.]

7. If the squares on two sides of a triangle together are greater (in area) than the square on the third side then the angle contained by the two sides is less than a right angle.

NOTE.

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.

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.

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.

SECTION IX.

GEOMETRICAL DRAWING. (CONTINUED.)

PARALLELS AND PARALLELOGRAMS.

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.

A

E

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.

Proof.

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.

Q.E.F.

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.

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.

Then APPQ=QR=RS=SB.

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.

EXAMPLES LI.

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.