A B is parallel to CE (a).
AC falls upon them.
.. by Euc. I. 29,

(b) angle BAC is equal to the alternate angle A CE.

Proof.

Again, AB is parallel to CE (a).

BD falls upon them.

by Euc. I. 29,

(c) angle ABC is equal to exterior angle ECD.

But the two angles,

ACE in (b), and E CD in (c),

make up the whole angle ACD.

... the exterior angle ACD is equal to the two interior opposite angles BAC, ABC.

Q. E. D.

THEOREM (Euclid I. 32, Second Part).

Repeat.-The enunciation of Euc. I. 13 and Axioms 1 and 2a.

General Enunciation.

The three interior angles of every triangle are equal to two right angles.

Particular Enunciation.

Given. The triangle

Construction.

B

Produce BC to D (see next page).

(a) Through the point C draw CE parallel to AB.

I

Proof.

Angle A CD is equal to angles B A C, A B ̊C. (Conclusion of last theorem.) Add angle A CB to each of these equals.

N

B

Then, by Axiom 2a,

(b) The two angles A CD, ACB are equal to the three angles B A C, A B C, A C B. But, by Euc. I. 13,

Angles A CD, A CB are equal to two right angles. by Axiom I,

angles BAC, ACB, ABC are equal to two right angles.

Q. E. D.

EXERCISE.

isosceles

Given.-ABC an
triangle, having A B equal to
A C, AD equal to AB.
Required.-To prove

that
angle B CD is a right angle
(DAC is the exterior angle
of triangle ABC: show that
ADC is an isosceles triangle:
and use Euc. I. 32, Second
Part.)

B

D

THEOREM (Euclid I. 33).

Repeat. The enunciations of Euc. I. 4; I. 27; I. 29; and Axiom 2a.

General Enunciation.

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

Particular Enunciation.

Given. (a) The straight lines A B, CD, equal and parallel.

B

D

Construction.
Join CB.

(b) The straight line AC joining the extremities A and C, and

(c) The straight line BD joining the extremities B and D.

Required. To prove that AC and BD are equal and parallel.

(NOTE. A C is said to join the extremities towards the same parts. CB joins the extremities towards opposite parts.)

Proof.

AB is parallel to CD (Given a).
BC meets them.

.. by Euc. I. 29,

(d) the angle ABC is equal to the alternate angle BCD.

Again, AB is equal to CD (Given a),
to each of these add BC,
.. by Axiom 2a,

(e) the two sides A B, BC, are equal to the two sides D C, C B, each to each.

So that in the two triangles ABC, DCB,

we have two sides equal to two sides (e),
and the included angles equal (d).
.. by Euc. I. 4, we have

(ƒ) the base A C equal to the base B D, and
(g) the angle ACB equal to the angle DBC.
So, the straight line B C meets A C, B C,
and makes the alternate angles equal (g).
.. by Euc. I. 27,

AC is parallel to BC, and
AC is equal to BC (ƒ).

EXERCISE.

Given.-A B, CD parallel,
A C, BD equal.

B

Q. E. D.

Required. To prove that angles CAB, BDC are together equal to two right angles.

Foin CB. And use

(Draw B E parallel to A C. Euc. I. 33, I. 5, I. 8, I. 13, I. 32.)

DEFINITIONS.

A quadrilateral is a four-sided figure.

A square has four equal sides and four right angles.

A rectangle has four right angles, and its opposite sides equal; but its adjacent sides unequal.

A parallelogram is a four-sided figure having its opposite sides parallel.

(A square and a rectangle are parallelograms, so are Figures A and B.)

DA

A

A is a rhombus, B is a rhomboid.

A trapezium is any four-sided figure that is not a square, a rectangle, or a parallelogram, as C.

A diameter of a parallelogram is the straight line which joins any two of its opposite angles.