PROP. XXXII. THEOREM.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles.

Let ABC be a triangle, and let one of its sides, BC, be produced to D.

Then it is to be proved that

1. The exterior angle ACD

the two interior and

opposite angles CAB and ABC; and

2. The three interior angles of the triangle, ABC, BCA, and CAB = two right angles.

CONSTRUCTION.-Through the point C draw CE parallel to BA (I. 31.)

PROOF.-1. Because AB is parallel to CE (cons.), and AC falls upon them, therefore the alternate angle ACE = the alternate angle BAC (I. 29).

Also because AB is parallel to CE (cons.), and BD falls upon them, therefore the exterior angle ECD the interior and opposite angle ABC (I. 29).

We have therefore the angles ACE and ECD = the angles BAC and ABC. But the angles ACE and ECD make together the angle ACD, and therefore the angle ACD = the angles BAC and ABC (ax. 2).

Therefore, it is proved, as required, that

1. The exterior angle ACD

the two interior and

opposite angles, CAB and ABC.

2. Next. Because, as we have just proved, the angle ACD =the angles BAC and ABC, add to each of these equals the

angle ACB, and then the angles ACD and ACB = the angles CAB, ABC, and ACB (ax. 2).

But the angles ACD and ACB two right angles (I. 13), and therefore also the angles CAB, ABC, and BCA two right angles (ax. 1).

Therefore, it is proved, as required, that

2. The three interior angles of the triangle, ABC, BCA, and CAB two right angles.

Wherefore,

If a side of any triangle, &c.

Exercises.

Q. E. D.

1. If AD bisecting the vertical angle A of a triangle ABC, bisects also the base BC in D; prove that ABC is an isosceles triangle.

2. If ABC be a triangle right-angled at A, and AD be drawn to bisect the base BC in D; prove that AD or DC.

BD

3. If ABC be an isosceles triangle with vertex A, and the side BA be produced beyond A to D; prove that the exterior angle CAD twice the angle ABC or the angle ACB, the angles at the base.

Because any rectilineal figure, as ABCDE, can, by drawing straight lines from a point F, within the figure, to each angle, be divided into as many triangles as the figure has sides, as is shewn in the above figure, and because the three interior angles of every triangle equal two right angles (I. 32),

The angles at F= four right angles (I. 15, Cor. 2);

Therefore, it is proved, as required, that

All the interior, &c.

Q. E. D.

Because the interior angle ABC with its adjacent exterior two right angles (I. 13),

angle ABD

Therefore

All the interior angles of the figure with four right angles (ax. 1).

And taking away from each the interior angles common

to both,

Then

All the exterior angles of the figure four right angles (ax. 3).

Similarly it may be demonstrated for any other rectilineal figure.

Therefore, it is proved, as required, that

All the exterior angles of any = Four right angles. rectilineal figure

Q. E. D.

PPOP. XXXIII. THEOREM.

The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel.

Let AB and CD be two equal and parallel straight lines, joined towards the same parts by the straight lines AC and BD.

Then it is to be proved that

1. The straight lines AC and BD

ww

each other.

2. The straight lines AC and BD are parallel to each other.

A

D

CONSTRUCTION.-Join AD (post. 1).

PROOF.-1. Because AB is parallel to CD, and AD meets them (hyp.), therefore the alternate angle BAD = the alternate angle ADC (I. 29).

Next, because in the triangles BAD and ADC, we have the sides BA and AD, and their angle BAD, in the former = the sides CD and DA, and their angle CDA, in the latter, each to each (hyp. cons. and proof above), therefore the angle ADB = the angle DAC, and the base AC = the base BD (I. 4).

Therefore, it is proved, as required, that

1. The straight lines AC and BD

each other.