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.

Join DA and BD to divide the given figure into triangles, and produce | AB indefinitely. D M

N Through. E draw EH parallel to DA, and through .C, CF parallel to DB : join DH and DF;

E then the triangle DHF is equal in area to the

С given figure ABCDE.

Next, through. D draw DN parallel to HF; bisect HF, a side of the triangle DHF, in L; at. L raise the perpendicular LM, and through H A LB . F draw FN a parallel to L M; the fig. LMNF is a rectangle of the same altitude as the triangle, and on half its base,—and is therefore equal in area to the triangle DHF.

3. In a similar way a crooked boundary, ABCDE, between two fields, M and N, may be made straight without changing the relative size of the fields. Draw AC the subtend to angle B, and

Emvrea
through. B, a st. line BF parallel to AC, and
join ČF; the crooked boundary AB, BC is
now converted into the single boundary CF. In M

D
С

N
a similar way FC and CD will be converted into
one boundary, and this last and D E into a single

B boundary ; and thus the crooked boundaries A B, BC, CD, and DE, will be changed into one straight boundary without affecting the size of the fields : the shape will be altered, but the areas

A F of fields M and N remain as at first.

G

PROP. 46. -PROB.

To describe a square on a given straight line.

Sol.-P. 11. To draw a st. line at right angles to a given st. line from a

given point in the same.
P 3. From the greater of two st. lines to cut off a part equal to the less.
P. 31. To draw a st. line through a point parallel to a given st. line.
Def. A. A parallelogram is a four-sided figure of which the opposite sides

are equal and parallel, and the diagonals join opposite angles. DEM.-P. 34. The opposite sides and angles of parallelograms are equal.

Ax. 1. Things equal to the same, are equal to each other.
P. 29. If a line fall on two parallel st. lines, it makes the alternate

angles equal to each other; and the ext. angle equal to the int. and
opposite angle on the same side ; and the two int. angles equal to

two rt. angles. Ax. 3. If equals be taken from equals, the remainders are equal. Def. 30. Of four-sided figures a square is that which has all its sides

equal, and all its angles rt. angles.

Exp. 1 | Datum.

2 Quces.

Given the st. line AB;
it is required to draw a square

on AB.

AB;

=

Cons. 1 by P. 11. From . A draw A C at rt. angles

to AB; 2 P. 3. make AD

AL

JB 3 P. 31. through. D draw DE || AB;

and through . B, BE || AD;
4 Def. A. then ABDE is a parallelogram,
5 Sol.

and the square required.
Dem. 1 by P.34&C.2. :: AB = DE, AD BE, and AB

AD; 2 | Ax. 1.

.. AB

AD DE EB, 3 D. 1 & 2. and .. the OABD E is equilateral. 4 C. 3. Also, : AD meets the parallels AB and DE, 5 P. 29. :, the int. LS BAD, ADE two rt. angles : 6 C. 1.

but the BAD is a rt. angle ; 7 Ax. 3. :. _ ADE is also a rt. angle. 8 P. 34. Now the opposite angles of os are equal ; 9 D. 7 & 6. :. _ ABE opposite to Z ADE is a rt. angle,

and 1 BED opposite to 2 BAD is a rt. angle;
10 D. 6, 7, & 9. -. the figure ADEB is a rectangle ;
11 D. 3.

it is also equilateral ;
12 Def. 30. therefore a ABED is a square on AB.

Q.E. F. Cor. 1.The squares on equal st. lines are equal; and if the squares are equal, the st. lines are equal."

2. Every parallelogram having one right angle, has all its angles rt. angles.

Sch.—Given the diagonal AB to construct a square.
Cons.1 P. 11. At. A and. B draw perpendiculars

F
AE and BF;
2 P. 9. bisect the rt. angles by AC and

C

В B.
BC meeting in C ;
3 P. 31. through .s B and A draw BD ||
AC, and AD || CB;

E 4 Sol. then ABCD is the square required. DEM. 1 byC.2,&P.6. ::the zs CAB and C B A are equal,

side AC = side CB;
2 P. 34. and : ACBD is a O, AD = CB, and BD = CA;
3 Concl. :: AC = C B = BD = D A, and the fig. is equilateral.
4 C.

Again, the 2s CAB, CB A being together one rt, engle, 5 P. 32, the angle C is a rt. angle ;

K

Dem. 6 P.46, Cor.2. But in a , as ACBD, when one angle is a rt. angle, all

the angles are rt. angles ; 7 C. 3, Concl. :. also ACBD has its angles rt. angles ; 8 Concl. And therefore, the figure being equilateral and rectan

gular, ACBD is the square required. USE AND APP.—The Geometrical Square is an instrument by means of which, and of the property of similar triangles that the sides about the equal angles are proportional, the height of an inaccessible object can be ascertained, provided a measurement to the perpendicular from the object can be made. The edges of the square are each divided into 100 equal parts, and from one corner a plummet is suspended ; when the object is seen along one edge of the instrument, the plummet cuts another edge, and forms a triangle similar to the triangle formed by lines representing the perpendicular from the object, a parallel to the horizontal line at its base, and the hypotenuse, or distance from the point of observation to the object itself.

n

In the adjoining figures, | A B represents

C the horizon ; p D a parallel to the horizon ; DB the height of the instrument; CD the height of the object C above the parallel to the horizon ; sp the edge along which the object is to be seen ; sr, rn, and pn graduated edges each of 100 parts; and p the point of suspension for the plummet. From the place of observativn measure

D the distance p D, and the height of the instrument DB; direct the edge sp towards the object C, and note the number of parts in sr or in rn. 1°. When the plummet cuts А

IB 8r in o, the triangle pso is similar to the

С triangle CDp; and we have the proportion 80:88:: PD: CD; whence CD = $p.pD and CB = CD + D B. 2o. When the plummet cuts rn in o, the triangles on p and

D CDp are similar ; and we have the proportiou pn : no :: PD:DC; whence CD = no.pD ; and CB = CD + DB.

A pn

B For example, let p D = 60 ft. ; 80 = 50 eq. pts. ; and DB = 6 ft. ; required

; CB. Here 50 : 100 :: 6n : 120 = CD, and 120 + 6 = 126 ft. = CB.-Sec Tate's Geometry, pp. 49-51.

80

n

Prop. 47.—THEOR.—(Most Important.) In any right-angled triangle, the square which is described upon the side subtending, or opposite to, the right angle, is equal to the squares described upon the sides containing the right angle. Cons.—P. 46. To describe a square on a given straight line.

P. 31. Through a point to draw a st. line parallel to a given st. line.

Pst. 1. Any two points may be joined by a st. line. Dem.-Def. 30. A square has all its sides equal, and its angles rt. angles. P. 14. If at a point in a st. line two other st.. lines on the opposite

si les of it make the adjacent anzles together equal to two rt. angles,

these two st. lines shall be in one and the same st. line.
Ax. 1. Thinys equal to the same, are equal to each other.
Ax. 2. If equals be added to equals, the wholes are equal.
P. 4. If two triangles have each two sides and their included angle equal,

the triangles are in every other respect equal.
P. 41. If a parallelogram and a triangle be upon the same base and

between the same parallels, the parallelogram is double of the

triangle.

Ax. 6. Things double of the same, are equal to each other. EXP. 1 Нур.

Let ABC be a A

and BAC a rt.
angle;
F

ң
2 Concl.

then the square
on BC = the

K
squares on AB
and AC.

H

B

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Cons.1 by P. 46.

Draw on BC a
square BE; on
BA, a square
BG; and on AC,

D
a square HC;

LE
through. A draw

AL || BD or to CE;
join A D and FC; also A E and BK.

2 P. 31.

3 Pst. 1.

DEM. 1 by H&Def.30 : 28 BAC and B AG are each a rt. angle ; 2 P. 14. :: the lines AC, AG on opp. sides of A B make

the adj. LS two rt. angles, and :. CA is in the same st. line with AG.

=

3 H. Def. 30 Also AB is on the same st. line with AH.

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I

С

DEM. 41C.1, Def. 301 :: the Ls DBC

and FBA are
each a rt. angle; F

H
5 Ax 1. ..the 2 DBC=

the L FBA:
6 Add.
To each of the

K equals add the

B В
L ABC;
7 | Ax. 2. :. the whole

DBA the

whole L FBC: 8 C.1,&Def.30 Hence, .. AB = FB, B D = BC,

L E and – DBA = ¿FBC; 9 P. 4. .:. base A D = base FC, and A ABD= AFBC. 10 C. 2.

Now the BL and the A ABD are both

on the same base BD, and between the same

parallels BD and AL; 11 P. 41. .. the BL is double of the A ABD. 12 C. 2. Also the square GB and the A FBC are

both on the same base FB, and between the

same parallels FB and G C; 13 P. 41. .. the

or square. GB, is double of the

A FBC:
14 D. 9. But the A ABD = the A FBC;
15 Ax. 6. ... the

BL

GB. 16 C. 3. Also, after joining A E, BK, the CL =

the square HC: 17 Ax. 2.

:. the whole square BDEC = the two squares

GB and HC;
18 C. 1. Now the squares are BE on side BC,

BG on side BA, and CH on side CA;
19 Concl. .. the square on side BC

on sides BA and CA.
20 Recap
Therefore, in any right-angled-triangle, &c.

Q.E.D. CoR. I.--Hence, if the sides of a rt. angled triangle be given in numbers, its hypotenuse may be found : for, let the squares of the sides be added together, and the square root of their sum will be the hypotenuse. Suppose AB the base, AC the perpendicular, BC the hypotenuse; the formula for B C is,

BC

NAB+ AC?

the square

the two squares

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