To make a Triangle, fimilar to another Triangle *.

Let ABC be the given Triangle, and DE a Line given, on which to construct a Triangle; fimilar, and in a fimilar Pofition, i. e. a- A like fituated, to ABC.

At the Point D, of the given Line, make an Angle EDa, equal to CAB, of the given Triangle, by

Pr. 49

And, at the Point E, make the Angle DEb, equal to ABC.
Produce Da and Eb, meeting in F.

Poft. 3.

The Triangle DEF will be fimilar to ABC. Q E. F.

DEM. For, the three Angles are refpectively equal. C.5.10.1.
Therefore, the Sides are proportional.
Confequently, the Triangles are fimilar. - -

P. 4. 6. ·
Def. 1. 6.

APPL. This Problem is of great ufe in various Profeffions. For, by it we learn to take Altitudes and Distances, though ever fo inacceffible; or, the Surveyor takes his Bearings, and lays down a Plan of the Ground he furveys; by it, the Mariner plans the Courte in which the Ship ploughs the Ocean; and the Mechanic plans the Ground, on which he intends to build, &c.; in fhort, it is almoft of univerfal Ufe; which, tɔ. enumerate, is not neceffary in this place.

For fimilar Figures, fee Def. 1. of the 6th Book of Elements.

B

B

PROBLEM

XVI.

18. VI. Euclid.

To make a Figure, fimilar to any given right-lined

Figure.

Let ABCDE be the Figure given, and a e a Line given, on which to conftruct a Figure, fimilar, and alike fituated to the given Figure.

E

Draw the Diagonals AC, AD, BE, and CE.

On the given Line, ae, make the Triangle abe, fimilar to the Triangle ABE, in the original Figure, by the foregoing. Pr. 15alfo, make ace fimilar to ACE, & ade to ADE Join bc & cd, which compleats the Figure. abcde is fimilar to ABCDE.

Q. E. F.

aThis method is deduced from the foregoing Problem; for, all right-lined Figures are compofed of, or may be reduced into right-lined Triangles. It is demonitrable from 4. and 13. of 6. El. The Application of this Problem is evident.

It may be thus performed, when required bigger.

On AB, the given Line, defcribe a Pentagon, Abcde, congruous to the given one, by reducing the given Figure into Triangles (as above) and making Abc, Acd, and Ade, refpective ly equal to them. Pr. 14. e E Produce AE, and the Diagonals, Ac, Ad, indefinite. Draw BC, CD, and DE, parallel to the Sides bc, cd, and de, refpectively; cuting the Diagonals and Side A E, in the Points C, D, and E. ABCDE is fimilar to Abcde.

After the fame manner, any Poligon may be increased or diminished in any Proportion; demonftrable by 2. 6. El.

PRO

PROBLEM

XVII. 46.1. Euclid.

To make a Square on a given Line, AB.

On either extreme of AB, make a Right

A

B

Wh. the Angle D=B; conf. D is a Right Angle.-10. I.
But, the Angles A and C are alfo equal; conf. Right.

Therefore, ABCD is a Square.

Def. 35.

To make a Rectangle; two Lines is given; AB & E.

E

B

The Square and Rectangle are of great ufe in the mechanic Arts; as moft regular Figure, are Rectangles.

In Menfuration, the Area of every Figure is reduced to a ftandard meafure; by the Square, or other Rectangle.

PRO

E

D

H

E

To conftruct a Parallelogram under a given Angle, and having its Sides equal to given Lines, of determinate length.

X

AB and E are the given Lines, and X the given Angle. At either extreme, of either Line (as A) make an Angle, BAD equal to the given Angle. Make AD equal to E, the other given Line. Through the Points D and B, draw DC parallel to AB, and BC to AD, meeting in C.-Pr. 5. Then is ABCD the Parallelogram required.

B

This needs no Demonftration, every thing being as required by Construction.

The oppofite Sides and Angles are equal, by

P. 15. 1. APPL. By this Problem is delineated Plans, &c. of four fided Objects which are not right-angled, but, whofe oppofite Sides are equal. Any one Angle, being taken, determines all the reft.

PROBLEM

XX. 42.I.Euclid.

To make a Rectangle, or any other angled Parallelogram, equal to a given Triangle.

BGG F

K D

Draw DG, from the point of bifection, per

pendicular to the Bafe AC, cuting EF in G. Laftly, draw CF parallel to DG.

The Rect. DGFC is equal to the Triangle ABC. QE F.

Or,

Or, if the Rectangle AHIC be conftructed on the whole Bafe, AC, and half the perpendicular Altitude, B K, it will be equal to the Triangle ABC.

DEM. For, it is half the Rectangle A EFC, which is double the Area of the Triangle ABC.

17. I.

N. B. If any other angled Parallelogram was required, you muft proceed as in Prob. 19. making the Angle CDG or DCF equal to the Angle given; ftill keeping the fame Altitude; i. e. between the fame Parallels, AC and EF, and the fame Bafe DC or AD.

SCHOL. From hence, and from the 17th Propofition of the firft Book of Elements, the whole Theory of Menfuration is deduced; as all Figures whatever, except Parallelograms, are refolved into Triangles in Menfuration.

From the 17th of the firft Book we learn, that every Triangle is equal to half a Parallelogram of the fame Baje and Height; conjequently, the Rectangle DG FC is equal to the Triangle ABC, which is on half its Bafe, AC, and the fame height, GD, equal BK.

COR. Hence, the Rule for meafuring a Triangle is to multiply the Perpendicular, BK (equal GD) by half the Bafe, AC (equal DC) or the whole Bafe, AC, by half the Perpendicular BK (equal AH); the first gives the Rectangle DGFC, the other is the Rectangle AHIC; either of which is equal to the Triangle ABC.

For, if the whole Bafe AC be multiplied by the Perpendi cular BK, it gives the Area of the Rectangle AEFC which is double of the Triangle (17. 1,) confequently, half that Sum is the Area of the Triangle ABC; and alfo of the Rectangle DGFC or AHIC; which is, therefore, equal to the Triangle given.

PRO