PROPOSITION X., PROBLEM 5.

To bisect a given finite straight line.

Let AB be the given straight line:

It is required to divide it into two equal parts.

On AB describe an equilat. A ABC (I. 1). Bisect ACB by the straight line CD, meeting AB in D (I. 9).

AB shall be bisected at the point D.
In the As ACD, BCD

CD is common to the two As,

▲ ACD = BCD (cons.);

.. base AD = base DB (I. 4).

Wherefore the given straight line AB is bisected at the

point D.-Q.E.F.

PROPOSITION XI., PROBLEM 6.

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 the given point

F

Join CF.

E

in it.

It is required to draw from Ca straight line at right angles to AB. Take any point D in AC, from CB cut off CE = CD.

On DE describe the equilat. A DFE.

The straight line CF drawn from point C shall be at right angles to AB.

But when a straight line falling upon another straight line makes the adjacent angles equal, each of the angles is a right angle (def. 10).

.. each of the angles DCF, ECF is a right angle.

Wherefore from the given point C in AB, CF has been drawn at right angles to AB.-Q.E.D.

PROPOSITION XII., PROBLEM 7.

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 line of unlimited length, and C the given point without it.

AB.

It is required to draw from point C a perpendicular to

.. ▲ EGC = ≤ CGF, and they are adjacent angles. .. they are right angles, and CG is perpendicular to AB. Wherefore from point C, &c.-Q.E.D.

PROPOSITION XIII., THEOREM 6.

The angles which one straight line makes with another, on one side of it, either are, or are together equal to, two right angles. Let the straight line AB make with the straight line CD, on one side of it, the s

CBA, ABD.

These either are, or are together equal to, two rt. s. For if CBA=▲ DBA, each of them is a rt. L (def. 10).

E

A

But if not, from point B draw BE at right angles to CD (I. 11).

Then the 8 CBE, EBD = two rt. angles.

Now, the

CBE the two 2s CBA, ABE, to each of

=

these equals add the angle EBD.

.. the s CBE, EBD = the three angles CBA, ABE, EBD (ax. 2).

Again, ▲ DBA = the two s DBE, EBA, to each of these equals add ▲ ABC.

.. the angles DBA, ABC = the same three s DBE, EBA, ABC (ax. 3).

.. the s DBA, ABC = ≤s CBE, EBD = two rt. ≤s. Wherefore the angles, &c.-Q.E.D.

PROPOSITION XIV., THEOREM 7.

If at a point in a straight line two other straight lines, on 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 point B, in the straight line AB, let the two straight lines BC, BD on the opposite sides of AB make the adjacent 4s ABC, ABD together equal to two rt. s; BD shall be in the same straight line with CB.

For if not, let BE be in the same straight line with BC.
Then s ABC, ABE together two rt. s (I. 13),

From these equals take away the common angle ABC.

A

D

E

Then ▲ ABE = ▲ ABD, the less to the greater, which is impossible.

.. BE is not in the same straight line with BC, and in the same manner it can be shown that no other line but BD can be.

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

PROPOSITION XV., THEOREM 8.

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 at

point E. The and the

For

=

AEC shall be equal to ▲ DEB,
CEB to the AED.

AE meets CD, Zs AEC, AED

together two rt. s (I. 13).

And ⚫ DE meets AB, ▲s BED, AED

together two rt. s (I. 13).

..s AEC, AED togethers BED, AED.

From these equals take away the common

the remaining CEA remaining ≤ =

DEB.

In the same manner it may be shown that the CEB = LAED.

Wherefore if two straight lines, &c.-Q.E.D.

Corollary.-1. Hence it is manifest that if two straight lines cut one another the angles which they make at the point of section are together equal to four right angles. Cor. 2. And, consequently, that all the angles made by any number of straight lines meeting at one point are together equal to four right angles.

PROPOSITION XVI., THEOREM 9.

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

Let BC, a side of ▲ ABC, be produced to D.

ACD shall be greater than

The

▲ BEA = < FEC they are opposite verticals (I. 15). .. 4 ECF = ▲ EAB.

But ACD is greater than

.. Z ACD is greater than

ECF (ax. 9).

EAB, i.e. ▲ CAB.

In the same manner, if AC be produced to G it may be

shown that

B

BCG is greater than

ABC.

ACD is greater than ABC.

Wherefore if one side, &c.-Q.E.D.

PROPOSITION XVII., THEOREM 10.

Any two angles of a triangle are together less than two right

Let ABC be a triangle.

A

and opposite

D

C

ABC.

angles.

Any two of its angles are together less than two rt. s.

Produce BC to D.

Then

А АВС.

To each of these add

ACD is the exterior of

▲ ACD is greater than the interior

ACB.

Then s ACB, ACD are together greater than 8 ACB, ABC.

But s ACB, ACD together two rt. s (I. 13).

.. 2 ACB, ABC are together less than two rt. s.

In the same manner it may be shown that the angles BAC, ACB, as also s CAB, ABC are together less than two rt. 2s.

Wherefore any two angles, &c.-Q.E.D.

PROPOSITION XVIII., THEOREM 11.

The greater side of every triangle has the greater angle
opposite to it.

In A ABC let the side AC be greater than the side AB.
ABC shall be greater than the ACB.

The