Take any point D on the other side of AB; with C as centre, and at the distance CD describe a circle; let E and F be the points in which the circle intersects AB; bisect EF in the point G; (prop. 10) join CE, CG, CF.

Then the straight line CG, drawn from C, shall be
perpendicular to AB.

Demonstration.

Because, in the triangles CGE, CGF,
EG is equal to FG, (constr.)
and GC is common to the triangles,

and the base EC is equal to the base FC; (def. 11) therefore the angle EGC is equal to the angle FGC; (prop. 8)

and they are adjacent angles;

but "when a straight line standing on another straight line makes the adjacent augles equal to one another, the straight line which stands on the other is perpendicular to it." (def. 9)

Therefore a straight line CG has been drawn perpendicular to AB from the point C.

Q.E.F.

Exercises.

1. Given a straight line AB and a point C on one side of it; find a point D on the other side of it such that any point on AB shall be equidistant from C and D.

2. Find a point in a given straight line such that the angle formed by joining it to two given points on opposite sides of the line shall be bisected by the straight line.

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.

ABD ;

Let AB make with CD upon one side of it the angles ABC, then shall these be either two right angles, or together equal to two

let the angle ABC be equal to the angle ABD;
then each of them is a right angle, (def. 9)
that is, they are two right angles.

Secondly.-As in figure 2,

let the angle ABC be not equal to the angle ABD.
From B draw BE at right angles to CD; (prop. 11)

then the angles EBC, EBD are two right angles.

Now whatever angle EBA is added to EBC to make it become ABC, the same angle is taken off from EBD to make it become ABD; therefore ABC, ABD together are equal to EBC, EBD together: but EBC, EBD are two right angles;

therefore the angles ABC, ABD are together equal to two right angles.

Q.E.D.

Exercises.

1. If in fig. 1, AB be produced to F, shew that the angles at B are four right angles.

2. If two straight lines intersect each other at any point, the four angles are together equal to four right angles.

3. If any number of straight lines pass through a point, the angles which they form at that point are together equal to four right angles.

4. If an angle BAC be bisected by AE, and if CA be produced to any point D, and the angle BAD be bisected by AF; shew that EAF is a right angle.

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, 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 on opposite sides of AB make the adjacent angles ABC, ABD together equal to two right angles;

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

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

Demonstration.

Then because AB meets the straight line CBE, therefore the angles ABC, ABE are together equal to two right angles; (prop. 13)

but the angles ABC, ABD are together equal to two right angles;

(hyp.) therefore ABC, ABE together are equal to ABC, ABD together;

(ax. 1) from each of these equals take away the common angle ABC, then the angle ABE is equal to the angle ABD; (ax. 3) that is, a less angle is equal to a greater angle; which is absurd : therefore BE is not in the same straight line with BC.

In the same way it may be shewn that no other line

but BD is in the same straight line with BC;

therefore BD is in the same straight line with BC.

Exercises.

Q.E.D.

1. How would this proof be affected if BE were drawn on the other side

of AB?

2. If at B, BF be drawn so that the angles DBA, DBF are together equal to two right angles; shew that ABF is a straight line.

3. If from a point E in a straight line AB, two straight lines EC, ED be drawn on opposite sides of AB, such that the angle AED is equal to the angle BEC, but the angle AEC greater than the angle BED; then CED cannot be a straight line.

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 each other in the

point E;

then shall the angle A EC be equal to the angle BED,

and the angle AED to the angle BEC.

C

A

E

B

D

Demonstration.

Because AE meets CD at the point E,

therefore the angles AEC, AED are together equal to two right angles (prop. 13).

:

also because DE meets AB at the point E,

therefore the angles BED, AED are together equal to two right angles. (prop. 13).

Hence the angles AEC, AED together are equal to the angles BED, AED together:

from each of these equals take away the common angle AED, then the angle AEC is equal to the angle BED. (ax. 3) Similarly it may be shewn that

the angle AED is equal to the angle BEC.

Exercises.

Q.E.D.

1. Write out the proof that the angle AED is equal to the angle BEC.

2. If two straight lines bisect each other, the straight lines joining their extremities form a figure of which the opposite sides are equal.

3. If two equal straight lines bisect each other, the straight lines joining their extremities are all equal.

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 of which one side BC is produced to D; then shall the exterior angle ACD be greater than either of the interior opposite angles BAC or ABC.

Bisect AC in E (prop. 10); join BE;

and produce BE to F, making EF equal to EB: (prop. 3).

join FC.

Demonstration.

Then in the triangles AEB, CEF,

(constr.) (prop. 15)

AE is equal to CE, and BE is equal to FE, and the angle AEB is equal to the angle CEF; therefore the angle EAB is equal to the angle ECF: (prop. 4) but the angle ECD is greater than the angle ECF, (ax. 9) therefore the angle ECD is also greater that the angle EAB,

that is, the angle ACD is greater than the angle CAB. Similarly, if 4C be produced to G, it may be shewn that the angle BCG is greater than the angle ABC,

but the angle BCG is equal to the angle ACD, (prop. 15) therefore the angle ACD is greater than the angle ABC.

Exercises.

Q.E.D.

1. Shew that the angle BEC is greater than the angle ABE; also that the

angle FCD is greater than the angle ABE.

2. Why is it necessary to have the word opposite in the enunciation ?