the two sides DC, CE, and their contained angles equal, are identical (th. 1), and therefore have the side AE equal to EB.

PROBLEM AV.

At a given point C, in a line AB, to erect a perpendicular. From the given point C, with any radius, cut off any equal parts CD, CE, of the given line; and, from the two centres D and E, with any one radius, describe arcs intersecting in F; then join CF, which will be perpendicular as required.

AD C EB

Drew the two equal radii DF, EF. Then the two triangles CDF, CEF, having the two sides CD, DF, equal to the two CE, EF, and CF common, are mutually equilateral; consequently they are also mutually aquiangular (th. 5), and have the two adjacent angles at C equal to each other; therefore, the line CF is perpendicular to AB (def. 11).

OTHERWISE.

When the point C is near the end of the line. From any point D, assumed above the line, as a centre, through the given point C describe a circle, cutting the given line at E; and through E and the centre D), draw the diameter EDF; then join CF, which will be the perpendicular required.

E

CB

For the angle at C, being an angle in a semicircle, is a right angle, and therefore the line CF is a perpendicular (by def. 15).

PROBLEM V.

From a given point A, to let fall a perpendicular on a given line BC.

From the given point A as a centre, with any convenient radius, describe an arc, cutting the given line at the two points D and E; and from the two centres D, E, with any radius, describe two arcs, intersecting at F; then draw AGF, which will be perpendicular to BC as required.

B

Draw the equal radii AD, AE, and DF, EF. Then the two triangles ADF, AEF, having the two sides AD, DF, equal to the two AE, EF, and AF, common, are mutually equilateral; consequently, they are also mutually equiangular (th. 5), and have the angle DAG equal the angle EAG Hence then, the two triangles ADG, AEG, having the two sides AD, AG, equal to the two AE, AG, and their included angles equal, are therefore equiangular (th. 1), and have the angles at G equal; consequently AG is perpendicular to BC (def. 11).

OTHERWISE.

When the point is nearly opposite the end of the line.

From any point D, in the given line BC, as a centre, describe the arc of a circle through the given point A, cutting BC in E; and from the centre E, with the radius EA, describe another arc, cutting the former in F; then draw AGF, which will be perpendicular to BC as required.

Draw the equal radii DA, DF, and EA, EF. Then the two triangles DAE, DFE, will be mutually equilater,

B

EC

consequently, they

are also mutually equiangular (th. 5), and have the angles at D equal. Hence, the two triangles DAG, DFG, having the two sides DA, DG, equal to the two DF, DG, and the included angles at D equal, have also the angles at G equal (th. 1); consequently, those angles at G are right angles, and the line AG is perpendicular to DG.

arc.

PROBLEM VI.

To make a triangle with three given lines AB, AC, BC.

With the centre A, and distance AC, describe an With the centre B, and distance BC, describe another arc, cutting the former in C. Draw AC, BC, and ABC will be the triangle required.

For the radii, or sides of the triangle, AC, BC, are equal to the given lines AC, BC, by construction.

A

B

A

B.

Note. If any two of the lines are not together greater than the third, the construction is impossible.

PROBLEM VII.

At a given point A, in a line AB, to make an angle equal to a given angle C.

From the centres A and C, with any one radius, describe the arcs DE, FG. Then, with radius DE, and centre F, describe an arc, cutting FG in G. Through G draw the line AG, and it will form the angle required.

Conceive the equal lines or radii, DE, FG, to be drawn. Then the two triangles CDE, AFG, being

E

FB

mutually equilateral, are mutually equiangular (th. 5), and have the angle at A equal to the angle at C

PROBLEM VIII.

Through a given point A, to draw a line parallel to a given line BC.

From the given point A draw a line AD to any point in the given line BC. Then draw the line EAF making the angle at A equal to the angle at D (by prob. 5); so shall EF be parallel to BC as required.

EA

DC

For, the angle D being equal to the alternate angle A, the lines BC, EF, are parallel, by th. 13.

PROBLEM IX.

To divide a line AB into any proposed number of equal parts.

Draw any other line AC, forming any angle with the given line AB; on which set off as many of any equal parts AD, DE, EF, FC, as the line AB is to be divided into. Join BC; parallel to which draw the other lines FG, EH, DI: then these will divide AB in the manner required. For those parallel lines divide both the sides AB, AC, proportionally, by th. 82.

A

F

E

PROBLEM X.

To make a square on a given line AB.

Raise AD, BC, each perpendicular and equal to AB; and join DC: so shall ABCD be the square sought.

D

A

For all the three sides AB, AD, BC, are equal, by the construction, and DC is equal and parallel to AB (by th. 24); so that all the four sides are equal, and the opposite ones are parallel. Again, the angle A or B, of the parallelogram, being a right angle, the angles are all right ones (cor. 1, th. 22). Hence, then, the figure, having all its sides equal, and all its angles right, is a square (def. 34).

PROBLEM XI.

To make a rectangle, or a parallelogram, of a given length and breadth, AB, BC.

Erect AD, BC, perpendicular to AB, and each equal

to BC; then join DC, and it is done.

The demonstration is the same as the last problem. And in the same manner is described any oblique parallelogram, only drawing AD and BC to make the given oblique angle with AB, instead of perpendicular to it.

D

A

B

B.

-C

PROBLEM XII.

To make a rectangle equal to a given triangle ABC.

Bisect the base AB in D; then raise DE and BF perpendicular to AB, and meeting CF parallel to AB, at E and F; so shall DF be the rectangle equal to the given triangle ABC (by cor. 2, th. 26).

CE

PROBLEM XIII.

A

To make a square equal to the sum of two or more given squares, Let AB and AC be the sides of two given squares. Draw two indefinite lines AP, AQ at right angles to each other; in which place the sides AB, AC, of the given squares; join BC: then a square described on BC will be equal to the sum of the two squares described on AB and AC (th. 34).

In the same manner, a square may be made equal to the sum of three or more given squares. For, if

A

A

-BE

D

AB, AC, AD be taken as the sides of the given squares, then, making AE= BC, AD = AD, and drawing DE, it is evident that the square on DE will be equal to the sum of the three squares on AB, AC, AD. And so on for more squares.

PROBLEM XIV.

To make a square equal to the difference of two given squares, Let AB and AC, taken in the same straight line, be equal to the sides of the two given squares. From the centre A, with the distance AB, describe a circle⚫ and make CD perpendicular to AB, meeting the circumference in D: so shall a square described on CD be equal to AD AC2, or AB2. AC3, as required (cor. th. 34).

PROBLEM XV.

To make a triangle equal to a given quadrilateral ABCD, Draw the diagonal AC, and parallel to it DE, meeting BA produced at E, and join CE; then will the triangle CEB be equal to the given quadrilateral ABCD.

For, the two triangles ACE, ACD, being on the same base AC, and between the same parallels AC, DE, are equal (th. 25); therefore, if ABC be added to each, it will make BCE equal to ABCD (ax. 2).

D

CB

PROBLEM XVI.

To make a triangle equal to a given pentagon ABCDE Draw DA and DB, and also EF, CG, parallel to them, meeting AB produced at F and G; then draw DF and DG; so shall the triangle DFG be equal to the given pentagon ABCDE.

For, the triangle DFA = DEA, and the triangle DGB = DCB (th. 25); therefore, by adding DAB to the equals, the sums are equal (ax. 2), that is, DAB

E

D

+DAF+DBG = DAB+ DAE + DBC, or the triangle DFG = to the pentagon ABCDE.

PROBLEM XVII.

D

G

To make a square equal to a given rectangle ABCD Produce one side AB, till BE be equal to the other side BC. On AE as a diameter describe a circle meeting BC produced at F: then will BF be the side of the square BFGH, equal to the given rectangle BD, as required; as appears by cor. th. 87, and th. 77.

PROBLEM XVIII.

To describe a circle about a given triangle ABC.
Bisect any two sides with two of the perpendiculars
DE, DF, DG, and D will be the centre.

For, Join DA, DB, DC. Then the two right-angled triangles DAF, DBE, have the two sides, De, ea, equal to the two DE, EB, and the included angles at E equal: these two triangles are therefore identical (th. 1), and have the side DA equal to DB. In like manner it is shown, that DC is also equal to DA or DB.

A

E

So that all the three, DA, DB, DC, being equal, they are radii of a circle passing through A, B, and C.

Note. The problem is the same in effect when it is required—

To describe the circumference of a circle through three given points A, B, C.

Then, from the middle point B draw chords BA, BC, to the two other points, and bisect these chords perpendicularly by lines meeting in O, which will be the centre. Again, from the centre O, at the distance of any one of the points, as OA, describe a circle, and it will pass through the two other points, B, C, as required. 'The demonstration is evidently as above.

PROBLEM XIX.

An isosceles triangle ABC being given, to describe another on the same base AB, whose vertical angle shall be only half the vertical angle C.

From C as a centre, with the distance CA, describe the circle ABE. Bisect AB in D, join DC, and produce to the circumference E, join EA and EB, and ABE shall be the isosceles triangle required.

For, since in the triangle EDA, EDB, `AD is equal to DB, and DE common to both, and the right angle EDA, equal to the right angle EDB, the side EA must be equal to the side EB, the tri

angle AEB, is therefore isosceles, and the angle

D

B

ACB at the centre, must be double of the angle AEB at the circumference

for they both stand on the same segment AB.