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BOOK I. that the parallelogram CL is equal to the square HC; Therefore the whole square BDEC is equal to the two squares GB, HC;

and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC:

Wherefore the square upon the side BC is equal to the squares upon the sides BA, AC.

Therefore, in any right-angled triangle, &c. Q. E. D.

a

b

xi 1.

PROP. XLVIII. THEOR.

If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it; the angle contained by these two sides is a right angle.

If the square described upon BC,

one of the sides of the triangle ABC,

be equal to the squares upon the other side BA, AC,
the angle BAC is a right angle.

From the point A draw a AD at right angles to AC, and make AD equal to BA,

and join DC.

Then, because DA is equal to AB,

the square of DA is equal to the

square of AB.

To each of these add

the square of AC;

therefore the squares of DA, AC

B

are equal to the squares of BA, AC.

xlvii. 1. But the square of DC is equal to the squares of DA, AC,

because DAC is a right angle;

By Hyp. and the square of BC is equal to the squares of BA, AC;

therefore the square of DC is equal to the square of BC;

and therefore also the side DC is equal to the side BC.
And because the side DA is equal to AB,

and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC; and the base DC is equal to the base BC;

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therefore the angle DAC is equal to the angle BAC; but DAC is a right angle;

therefore also BAC is a right angle.

Therefore, if the square, &c. Q. E. D.

[The easiest mode of drawing this figure is by producing the line BA to D. It must be remembered, that though AD does lie in the same straight line with AB, this is in the present case only a consequence of its being drawn at right angles to AC.]

BOOK I.

• iii. 1.

BOOK I.

DEDUCTIONS FROM BOOK I.

1. To prove that the perpendicular is
the shortest straight line from a
given point to a given straight line.
Take any other straight line PR,
and prove by means of Propositions,
I. 32. and I. 19.

P

R

2. Through a given point to draw a straight line, which shall cut off equal segments from two given intersecting straight lines.

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4. To find the point in a given straight line, from which two BOOK I.

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5. Upon a given straight line to construct an isosceles right

angled triangle, whose right angle shall be subtended by the given line.

Bisect AB in O.

Draw OC at right

angles to it, equal to OA

or OB.

Join CA, CB.

Prove AC equal to A

CB by I. 4.

B

Then prove ACB to be a right angle by I. 32., showing that A and B are each half a right angle.

6. From two given points on the same side of a given straight line, to draw two straight lines which shall meet in it, and make equal angles with it.

From either point as P drop

a perpendicular, and produce it
to O, making AO equal to AP.
Join OQ. The lines PR,

RQ shall be those required.
Prove by I. 4., the angle PRA

Р

A

R.

B

equal to the angle ORA, and equal to QRB by means of

I. 15.

G

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10. The straight line which joins the points of bisection of

the sides of a triangle

is parallel to the base. Prove that the triangles ADB and AEB are each half of the given triangle

by I. 38.; and therefore

equal to each other, and A

D

therefore that DE is parallel to AB, by I. 39.

B

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