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PROP. V. PROB.

To bisect a given finite straight line, that is, to divide it into two equal parts.

Let AB be the given straight line ; it is required to divide it into two equal parts.

Describe (Prob. 1.) upon it an equilateral triangle ABC, and bisect (Prob. 4.) the angle ACB by the straight line CD. AB is cut into two equal parts in the

c

point D.

Because AC is equal to CB, and CD common to the two triangles ACD, BCD: the two sides AC, CD, are equal to the two BC, CD, each to each ; but the angle ACD is also equal to the angle BCD; therefore the base AD is equal to the base (1. 1.) DB, and the straight line AB is divided into two equal parts in the A point D.

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PROP. VI. PROB.

To draw a straight line at right angles to a given straight line, from a given

point in that line. Let AB be a given straight line, and C a point given in it; it is required to draw a straight line from the point C at right angles to AB.

Take any point D in AC, and (Prob. 3.) make CE equal to CD, and upon DE describe (Prob. 1.) the equilateral triangle DFE, and join FC;

F
the straight line FC, drawn from the
given point C, is at right angles to the
given straight line AB.

Because DC is equal to CE, and
FC common to the two triangles
DCF, ECF, the two sides DC, CF
are equal to the two EC, CF, each A D

C E B to each ; but the base DF is also equal to the base EF; therefore the angle DCF is equal (5. 1.) to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called a right (7. def.) angle; therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB, FC has been drawn at right angles to AB.

PROP. VII. PROB.

To draw a straight line perpendicular to a given straight line, of an un

limited length, from a given point without it. Let AB be a given straight line, which may be produced to any length both ways, and let C be a point without it. It is required to draw a straight line perpendicular to AB from the point C.

Take any point D upon the other side of AB, and from the centre C

C at the distance CD, describe (3. Post.) the circle EGF meeting AB in F, G: and biseçt (Prob. 5.) FG in H, and join CF, CH, CG; the straight line CH, drawn from the

A F H

G B given point C, is perpendicular to

D the given straight line AB.

Because FH is equal to HG, and HC common to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each; but the base CF is also equal (11. Def. 1.) to the base CG; therefore the angle CHF is equal (5. 1.) to the angle CHG; and they are adjacent angles; now when a straight line standing on a straight line makes the adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other is called a perpendicular to it; therefore from the given point C a perpendicular CH has been drawn to the given straight line AB.

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To construct a triangle of which the sides shall be equal to three given straight lines; but

any tuio whatever of these lines must be greater than the third. Let A, B, C be the three given straight lines, of which any two whatever are greater than the third, viz. A and B greater than C; A

K and C greater than B ; and B and C than A. It is required to make a triangle of which the sides shall be equal to A, B, C,

DH each to each.

T G Take a straight line DE, terminated at the point D, but unlimited towards E, and make (Prob. 3.) DF equal to A, FG

A to B, and GH equal to C; and from the centre F, at the distance

B FD, describe (3. Post.) the circle DKL; and from the centre

G, at the distance GH, describe (3. Post.) another circle HLK; and join KF, KG; the triangle KFG has its sides equal to the three straight lines, A, B, C.

Because the point F is the centre of the circle DKL, FD is equal (11. Def.) to FK; but FD is equal to the straight line A; therefore FK is equal to A : Again, because G is the centre of the circle LKH, GH is equal (11. Def.) to GK; but GH is equal to C; therefore, also, GK is equal to C; and FG is equal to B; therefore the three straight lines KF, FG, GK, are equal to the three A, B, C: And therefore the triangle KFG has its three sides KF, FG, GK equal to the three given straight lines, A, B, C.

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SCHOLIUM.

If one of the sides were greater than the sum of the other two, the arcs would not intersect each other : but the solution will always be possible, when the sum of two sides, any how taken (13. 1.) is greater than the third.

PROP. IX. PROB.

At a given point in a given straight line, to make a rectilineal angle equal to

a given rectilineal angle. Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle; it is required to make an angle at the given point A in the given straight line AB, that shall be equal to the given rectilineal angle DCE.

Take in CD, CE any points D, E, and join DE; and make (Prob. 8.) the triangle AFG, the sides of which shall be equal to the three

G straight lines, CD, DE, CE, so that

E CD be equal to AF, CE to AG,

F and DE to FG; and because DC, D CE are equal to FA, AG, each to each, and the base DE to the base FG; the angle DCE is equal (5.

B 1.) to the angle FAG. Therefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE.

PROP. X. PROB.

A

Two angles of a triangle being given, to find the third. Draw any straight line CD; at a point therein, as B, make the angle CBA equal to one of the given angles, and the angle ABE equal to the other : the remaining angle EBD will be the third angle required; because those three angles (Cor. 6. 1.) are together equal to two right angles.

C

B

E

D

PROP. XI. PROB.

Two angles of a triangle and a side being given, to construct the triangle.

The two angles will either be both adjacent to the given side, or the one adjacent and the other opposite : in the latter case, find the third angle (Prob. 10.); and the two adjacent angles will thus be known. Draw the straight line BC equal to the

A given side ; at the point B, make an angle CBA equal to one of the adjacent angles, and at C, an angle BCA equal to the other; the two lines BA, CA, will intersect each other, and form with BC the triangle required; for if they were parallel, the angles B, C, would be together B equal to two right angles, and therefore could not belong to a triangle : hence, BAC will be the triangle required.

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Two sides and an angle opposite to one of them being given, to construct the

triangle. This Problem admits of two cases. First. When the given angle

А. is obtuse, make the angle BC'A equal to the given angle; and take C'A equal to that side which is adjacent to the given angle, the arc described from A as a centre, with a radius equal to AB, the other given side, would cut BC on opposite sides of C'; so that only one obtuse angled triangle could be

B formed; that is, the triangle BC'A will be the triangle required.

And, if the given angle were right, although two triangles would be forned, yet, as the hypotenuse would meet BC at equal distances from the common perpendicular, these triangles would be equal.

Secondly. If the given angle be acute, and the side opposite to it greater than the adjacent side, the same mode of construction will apply: for, making BCA equal to the given angle, and AC equal to the adjacent side ; then, from A as centre, with a radius equal to the other given side, describe an arc, cutting CB in B ; draw AB, and CAB will be the triangle required.

But if the given angle is acute, and the side opposite to it less than the other given side ; make the angle CBA equal to the given angle, and take BA equal to the adjacent side ; then, the arc described from the centre A, with the radius AC equal to the opposite side, will cut the straight line BC in two points C' and C, lying on the same side of B : hence, there will be two triangles BAC', BAČ, either of which will satisfy the conditions of the problem.

SCHOLIUM. In the last case, if the opposite side was equal to the perpendicular from the point A on the line BC, a right angled triangle would be formed. And the problem would be impossible in all cases, if the opposite side was less than the perpendicular let fall from the point A on the straight line BC.

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To draw a straight line through a given point parallel to a given straight line.

Let A be the given point, and BC the given straight line, it is required to draw a straight line through the point A, parallel to the straight line

A BC.

E In BC take any point D, and join AD; and at the point A, in the straight line AD, make (Prob. 9.) the angle DAE equal to the angle

B D

C ADC; and produce the straight line EA to F.

Because the straight line AD, which meets the two straight lines BC, DF, makes the alternate angles EAD, ADC equal to one another, EF is parallel (19. 1.) to BC. Therefore the straight line EAF is drawn through the given point A parallel to the given straight line BC.

PROP. XIV. PROB.

To describe a parallelogram that shall be equivalent to a given triangle, and have

one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.

Bisect (Prob. 5.) BC in E, join AE, and at the point E in the straight

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