To describe a square upon a given straight line.

Let AB be the given st. line.

It is required to describe a square on AB.
From A draw AC 1 to AB.

In AC make AD=AB.
Through D draw DE || to AB.
Through B draw BE || to AD.

Then AE is a parallelogram,

and.. AB=ED, and AD BE.

But AB= AD,

:. AB, BE, ED, DA are all equal;
.. AE is equilateral.

And BAD is a right angle,

.. AE is a square,

and it is described on AB.

Def. xxx.

Q. E. F.

Ex. 1. Shew how to construct a rectangle whose sides are equal to two given straight lines.

Ex. 2. Shew that the squares on equal straight lines are equal.

Ex. 3. Shew that equal squares must be on equal straight lines.

NOTE. The theorems in Ex. 2 and 3 are assumed by Euclid in the proof of Prop. XLVIII. See also Prop. A. of Book II.,

p. 77.

PROPOSITION XLVII. THEOREM.

In any right-angled triangle the square which is described on the side subtending the right angle is equal to the squares described on the sides which contain the right angle.

Let ABC be a right-angled ▲, having the rt. ≤ BAC.
Then must sq. on BC= sum of sqq. on BA, AC.

On BC, CA, AB descr. the sqq. BDEC, CKHA, AGFB. Through A draw AL || to BD or CE, and join AD, FC.

Then

and ::

BAC and L BAG are both rt. 4s,
.. CAG is a st. line;

LBAC and L CAH are both rt. 48;
.. BAH is a st. line.

Now DBC= ▲ FBA, each being a rt. 4,

I. 14.

I. 14.

Then in AS ABD, FBC,

.. Δ ABD= Δ FBC.

Now

:: AB=FB, and BD=BC, and ▲ ABD

LFBC,

I. 4.

BL is double of ▲ ABD, on same base BD and between same ||s AL, BD,

I. 41.

and sq. BG is double of ▲ FBC, on same base FB and between same s FB, GC;

I. 41.

BL=sq. BG.

Similarly, by joining AE, BK it may be shewn that
CL=sq. AK.

Now sq. on BC=sum of □ BL and □ CL,
=sum of sq. BG and sq. AK,

= sum of

sqq. on BA and AC.

Q. E. D.

Ex. 1. Prove that the square, described upon the diagonal of any given square, is equal to twice the given square.

Ex. 2. Find a line, the square on which shall be equal to the sum of the squares on three given straight lines.

Ex. 3. If one angle of a triangle be equal to the sum of the other two, and one of the sides containing this angle being divided into four equal parts, the other contains three of those parts; the remaining side of the triangle contains five such parts.

Ex. 4. The triangles ABC, DEF, having the angles ACB, DFE right angles, have also the sides AB, AC equal to DE, DF, each to each; shew that the triangles are equal in every respect.

NOTE. This Theorem has been already deduced as a Corollary from Prop. E, page 43.

Ex. 5. Divide a given straight line into two parts, so that the square on one part shall be double of the square on the other.

Ex. 6. If from one of the acute angles of a right-angled triangle a line be drawn to the opposite side, the squares on that side and on the line so drawn are together equal to the sum of the squares on the segment adjacent to the right angle and on the hypothenuse.

Ex. 7. In any triangle, if a line be drawn from the vertex at right angles to the base, the difference between the squares on the sides is equal to the difference between the squares on the segments of the base.

PROPOSITION XLVIII. THEOREM.

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 those sides is a right angle.

D

A

B

Let the sq. on BC, a side of ▲ ABC, be equal to the sum of the sqq. on AB, AC.

Then must L BAC be a rt. angle.

From pt. A draw AD 1 to AC.

Make AD AB, and join DC.

Then

:: AD=AB,

.. sq. on AD=sq. on AB;
add to each sq. on AC:

I. 46, Ex. 2.

then sum of sqq. on AD, AC= sum of sqq. on AB, AC.
But sq. on DC=sum of sqq. on AD, AC,
and sq. on BC= sum of sqq. on AB, AC;
.. sq. on DC=sq. on BC,

.. DC=BC.

Then in As ABC, ADC,

I. 47.

Hyp.

I. 46, Ex. 3.

:: AB= AD, and AC is common, and BC= DC,

and DAC is a rt. angle, by construction,

.. BAC is a rt. angle.

I. C.

Q. E. D.

BOOK II.

INTRODUCTORY REMARKS.

THE geometrical figure with which we are chiefly concerned in this book is the RECTANGLE. A rectangle is said to be contained by any two of its adjacent sides.

Thus if ABCD be a rectangle, it is said to be contained by AB, AD, or by any other pair of adjacent sides.

We shall use the abbreviation rect. AB, AD to express the words "the rectangle contained by AB, AD.”

We shall make frequent use of a Theorem (employed, but not demonstrated, by Euclid) which may be thus stated and proved:

If the adjacent sides of one rectangle be equal to the adjacent sides of another rectangle, each to each, the rectangles are equal in area.

Then must rect. ABCD=rect. EFGH.

For if the rect. EFGH be applied to the rect. ABCD, so

that EF coincides with AB,

then FG will fall on BC,

EFG = L ABC,

and G will coincide with C, :: BC=FG.

Similarly it may be shewn that H will coincide with D,

.. rect. EFGH coincides with and is .. equal to rect. ABCD.

Q. E. D.