PROP. 47. THEOR. In any right-angled triangle the square which is described on the side subtending the right angle is equal to the sum of the squares described on the sides which contain the right angle. Given A ABC hav

ing the right angle

K к
To prove sq. on BC=

sum of squares on

BA and AC.
Describe squares on

BC, BA, and AC
(I. 46). Join KC
and AD, and draw

AL || BD (I. 81).
:: _BAH is a right
angle (Def. 24), and

E L BAC is a rt. angle (given), :: HA and AC are in the same st. line (I. 14). Similarly, BA and AG are in the same st. line. :: LKBA = L.DBC (Def. 24, Ax. 10), add – ABC to

each; :: LKBC = L ABD, and KB BA, and BC = BD, :: AKBC

A ABD (I. 4). Sq. BH is double KBC, they are on the same base

KB, and between the same parallels KB, HC (1. 41).
For a like reason, I BL is double A ABD.
:: BH = BL, doubles of equal things are equal (Ax. 6).
By a similar proof, sq. AF = | CL,

:: BH + AF BE,
i.e., BA? + AC=BCQ.E.D.

There are many ways of demonstrating Prop. 47 besides the above. The following is an easy, interesting method, and has this advantage—that it lends itself readily to demonstration by superposition or to mechanical illustration.

Place any two squares, as GB, EC, side by side, so that their bases are in the same straight line. Take CH = AB, and produce BE, making EK also=AB. Join GH, HD, DK, and KG. Since CH = AB, .: AH

= BC =FK.
.: triangles GAH, GFK,

KED, and HČD can
be proved equal to one
another by 1. 4.

E .: KDGH is equi

lateral. :: LKGF = L AGH, to each add , FGH. .: LKGH = LFGA=a

rt. L,

.. KGHD is also rectangular.


B H .:. it is a square.

sqq. on DC and CH, which are sides of a rt.-angled A.

:: sq. on HD

Suppose the figure GACDEF cut out in cardboard and cut along GH and HD. Let AGAH be turned about the point G as about a hinge into the position of GFK, and A DCH about D into the position of DEK, and thus produce the

sq. KGHD out of the sqq. GB and EC. 143. The square on the diagonal of a square equals twice the square on one side. 144. Find a square equal to three times a given square.

145. When the two sides of a right-angled triangle are given, show how to find the hypotenuse, (1) geometrically, (2) arithmetically.

(146. By how much is it shorter to go from comer to corner across a field than round by the two sides, which are respectively 100 yds. and 75 yds. long, and are at right angles to one another.


If the square described on one side of a triangle be equal to the sum of the squares described on the other two sides of it, the angle contained by these two sides is a right angle. Given A ABC having sq. on BC

squares on BA and AC. To prove – BAС a right

angle. At A in AC draw AD I

AC (1.11); make AD = B AB (Post. 3). Join DC ::BA = AD, :: BA2 AD2; to each add AC?. :: BA? + ACP AD2 + AC2. But BA? + AC2 = BO2 (given), and ADP + AC? - DCP (I. 47); :: BC2 = DC,

DC. And also BA, AC = DA, AC, :: ABAC = ADAC (1.8),

:: _BAC _DAC. But LDAC is a rt. angle (cons.),

:: · BAC is a rt. angle. Q.E.D. 147. If the square on a diagonal of a parallelogram equal the sum of the squares on two adjacent sides, the figure will be a rectangle.

148. If the square on one side of a triangle be greater than, or less than, the sum of the squares on the other two sides, of what kind will be the angle contained by those sides

.: BC

I. On Analysis and Synthesis. When a skilled workman gets an order with full specifications for the construction of a particular machine, be first prepares the necessary pieces of apparatus, such as wheels, axles, bolts, etc., and proceeds to fit these together in accordance with his given directions. On the other hand, if he be required to make a machine of the construction of which he is ignorant, instead of working according to detailed directions, he might procure a specimen, take it to pieces in order to learn the principle of its construction, and thus learn how to put it together again, and also how to make others like it.

These two methods of procedure—(1) following given specifications to produce a certain result, (2) finding the method of construction by dissecting a specimen-are important scientific pro

processes, called (1) Synthesis, or the method of composition, (2) Analysis, or the method of decomposition. Synthesis is the method employed in Geometry in the ordinary propositions in the text. Analysis is frequently used to find the principle, as of a required construction, and is followed by the ordinary synthesis. As Analysis may be found useful to the pupil in the solution of exercises, a few examples of its application are appended. It may be premised, however, that there is no golden rule to the solving of exercises : a few directions and examples can be given; the rest must be left to the pupil's ingenuity and perseverance, which qualities geometrical exercises are eminently fitted to bring out, i.e., to educate. The pupil is not to be discouraged when he meets a difficulty: he should not puzzle too long at a time over it when the mind is wearied, but should take it up again and again when he is fresh, and so persevere till he conquers.

Examples of Analysis. 1. Through a given point to draw a straight line to cut off equal parts from two given intersecting lines. Given point P and lines


B AB and AC. To draw a line from P

F to cut off equal parts A

from AB and AC. Analysis.-Suppose the

problem effected, and that PDE is the required line; then, since AD= AE, ADE is an isosceles triangle, and we know from previous work (Ex. 5) that if we bisect _ A, the bisector will be 1 DE. Now that we have learned that the required line is I the bisector of

the 2 A, the following construction is suggested :Synthesis.-Bisect 2 BAC (1.9); draw from P a line

I AF (1. 31), cutting AB at D, and AC at E. Proof. : _DAF = EAF

LEAF, and L. AFD = L'AFE, and AF is common, ... ΔAFD

A AFE (1. 26); :. AD AE. Q.E.F.

2. Draw through the middle point of one side of a triangle a straight line parallel to the base, it shall bisect the other side. Given AD = DB, and DE ||

To prove AE EC.
Analysis. — Suppose AE

EC, then DE, besides being B
|| BC, = half BC (Ex. 100).

Bisect BC at F, and join EF
EF is =

and || BD (I. 33) = AD, and FC = DE,

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