PROPOSITION XI. PROBLEM.

To draw a straight line at right angles to a given straight line, from a given point in the same.

Let AB be the given straight line, and C'a given point in it.

It is required to draw a straight line from the point C at right angles to AB

In AC take any point D, and make CE equal to CD; (1. 3.) upon DE describe the equilateral triangle DEF (1. 1,) and join CF. Then CF drawn from the point C,shall be at right angles to AB. Because DC is equal to EC, and FC is common to the two triangles DCF, ECF;

the two sides DC, CF are equal to the two sides EC, CF, each to each; and the base DF is equal to the base EF; (constr.)

therefore the angle DCF is equal to the angle ECF: (1. 8.)
and these two angles are adjacent angles.

But when the two adjacent angles which one straight, line makes with another straight line, are equal to one another, each of them is called a right angle: (def. 10.)

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.

Q.E.F.

COR. By help of this problem, it may be demonstrated that two straight lines cannot have a common segment.

If it be possible, let the segment AB be common to the two straight lines ABC, ABD.

E

B

From the point B, draw BE at right angles to AB; (1. 11.)
then because ABC is a straight line,

therefore the angle ABE is equal to the angle EBC. (def. 10.) Similarly, because ABD is a straight line,

therefore the angle ABE is equal to the angle EBD;
but the angle ABE is equal to the angle EBC,

wherefore the angle EBD is equal to the angle EBC, (ax. 1.)
the less equal to the greater angle, which is impossible.
Therefore two straight lines cannot have a common segment.
PROPOSITION XII. PROBLEM.

To draw a straight line perpendicular to a given straight line of unlimited length, from a given point without it.

Let AB be the given straight line, which may be produced 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.

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Upon the other side of AB take any point D,

and from the center C, at the distance CD, describe the circle EGF meeting AB, produced if necessary, in Fand G: (post. 3.) bisect FG in H (1. 10.), and join CH.

Then the straight line CH drawn from the given point C, shall be perpendicular to the given straight line AB.

Join FC, and CG.

Because FH is equal to HG, (constr.)

and HC is common to the triangles FHC, GHC;

the two sides FH, HC, are equal to the two GH, HC, each to each; and the base CF is equal to the base CG; (def. 15.) therefore the angle FHC is equal to the angle GHC; (1. 8.) and these are adjacent angles.

But when a straight line standing on another 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. (def. 10.)

Therefore from the given point C, a perpendicular CH has been drawn to the given straight line AB.

Q.E.F.

PROPOSITION XIII. THEOREM.

The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles.

Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD.

Then these shall be either two right angles, or, shall be together, equal to two right angles.

For if the angle CBA be equal to the angle ABD,
each of them is a right angle. (def. 10.)

But if the angle CBA be not equal to the angle ABD, from the point B draw BE at right angles to CD. (1. 11.) Then the angles CBE, EBD are two right angles. (def. 10.)

And because the angle CBE is equal to the angles CBA, ABE, add the angle EBD to each of these equals;

therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. (ax. 2.)

Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC;

therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC.

But the angles CBE, EBD have been proved equal to the same three angles;

and things which are equal to the same thing are equal to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC; but the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are together equal to two right angles. (ax. 1.)

Wherefore, when a straight line, &c. Q.E.D.

PROPOSITION XIV. THEOREM.

If at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles; then these two straight lines shall be in one and the same straight line.

At the point B in the straight line AB, let the two straight lines BC, BD upon the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles.

Then BD shall be in the same straight line with BC.

For, if BD be not in the same straight line with BC,

if possible, let BE be in the same straight line with it.

Then because AB meets the straight line CBE;

therefore the adjacent angles CBA, ABE are equal to two right angles; (I. 13.)

but the angles CBA, ABD are equal to two right angles; (hyp.) therefore the angles CBA, ABE are equal to the angles CBA, ABD: (ax. 1.)

take away from these equals the common angle CBA, therefore the remaining angle ABE is equal to the remaining angle ABD; (ax. 3.)

the less angle equal to the greater, which is impossible:
therefore BE is not in the same straight line with BC.

And in the same manner it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the same straight line with BC.

Wherefore, if at a point, &c. Q.E.D.

PROPOSITION XV. THEOREM.

If two straight lines cut one another, the vertical, or opposite angles shall be equal.

Let the two straight lines AB, CD cut one another in the point E. Then the angle AEC shall be equal to the angle DEB, and the angle CEB to the angle AED.

Because the straight line AE makes with CD at the point E, the adjacent angles CEA, AED;

these angles are together equal to two right angles. (1. 13.) Again, because the straight line DE makes with AB at the point E, the adjacent angles AED, DEB;

these angles also are equal to two right angles; but the angles CEA, AED have been shewn to be equal to two right angles; wherefore the angles CEA, AED are equal to the angles AED, DEB: take away from each the common angle AED,

and the remaining angle CEA is equal to the remaining angle DEB. (ax. 3.)

In the same manner it may be demonstrated, that the angle CEB is equal to the angle AED.

Therefore, if two straight lines cut one another, &c. Q. E. D.

COR. 1. From this it is manifest, that, if two straight lines cut each other, the angles which they make at the point where they cut, are together equal to four right angles.

COR. 2. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles.

PROPOSITION XVI. THEOREM.

If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

Let ABC be a triangle, and let the side BC be produced to D. Then the exterior angle ACD shall be greater than either of the interior opposite angles ČBA or BAC.

Bisect AC in E, (1. 10.) and join BE; produce BE to F, making EF equal to BE, (1. 3.) and join FC.

Because AE is equal to EC, and BE to EF; (constr.)

the two sides AE, EB are equal to the two CE, EF, each to each, in the triangles ABE, CFE;

and the angle AEB is equal to the angle CEF, because they are opposite vertical angles; (I. 15.) therefore the base AB is equal to the base CF, (1. 4.) and the triangle AEB to the triangle CEF,

and the remaining angles of one triangle to the remaining angles of the other, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF;

but the angle ECD or ACD is greater than the angle ECF; therefore the angle ACD is greater than the angle BAE or BAC. In the same manner, if the side BC be bisected, and AC be produced to G; it may be demonstrated that the angle BCG, that is, the angle ACD, (1. 15.) is greater than the angle ABC.

Therefore, if one side of a triangle, &c. Q.E.D.

Then

Any two angles of a triangle are together less than two right angles.
Let ABC be any triangle.

any two of its angles together shall be less than two right angles.

A

Produce any side BC to D.

Then because ACD is the exterior angle of the triangle ABC; therefore the angle ACD is greater than the interior and opposite angle ABC; (1. 16.)

to each of these unequals add the angle ACB;

therefore the angles ACD, ACB are greater than the angles ABC, ACB;

but the angles ACD, ACB are equal to two right angles; (1. 13.) therefore the angles ABC, ACB are less than two right angles.

In like manner it may be demonstrated,

that the angles BAC, ACB are less than two right angles,
as also the angles CAB, ABC.

Therefore any two angles of a triangle, &c. Q. E.D.

The greater side of every triangle is opposite to the greater angle.

Let ABC be a triangle, of which the side AC is greater than the side AB.