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Proof.

A B is parallel to CE (a).
A C falls upon them.

by Euc. I. 29, (6) angle BAC is equal to the alternate

angle ACE. Again, A B is parallel to CE (a). BD falls upon them.

by Euc. I. 29, (1) angle ABC is equal to exterior angle

ECD.
But the two angles,

ACE in (b), and ECD in (c),

make up the whole angle A CD. .. the exterior angle ACD is equal to the two interior opposite angles B A C, A BC.

Q. E. D.

THEOREM (Euclid 1. 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

A B C.
Required. – To prove

that the three angles
ABC, BCA, CAB are
together equal to two

right angles. Construction.

Produce B C to D (see next page).
(a) Through the point C draw CE parallel to A B.

B

I

Proof.
Angle A CD is equal to angles B AC, A B.C.

(Conclusion of last theorem.) Add angle A CB to each of these equals.

A

B

D

Then, by Axiom 2a, (6) The two angles A CD, ACB are equal to the three angles BAC, ABC, ACB.

But, by Euc. I. 13, Angles A CD, ACB are equal to two right angles.

by Axiom 1, angles BAC, ACB, A B C are equal to two right angles.

Q. E. D.

D

EXERCISE.
Given.-ABC an isosceles

triangle, having A B equal to

A C, A D equal to A B. Required.–To prove that angle BCD is a right angle (D AC is the exterior angle of triangle ABC: show that ADC is an isosceles triangle : and use Euc. I. 32, Second Part.)

A

B

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 extremi

ties of two equal and parallel straight lines towards the same parts, are them

selves equal and parallel. Particular Enunciation. Given.-(a) The straight lines A B, CD, equal

and parallel.

B

(6) The straight line AC joining the extremities

A and C, and (c) The straight line B D joining the extremities

B and D. Required.—To prove that A C and B D are

equal and parallel.

A

B

Construction.

Join CB.

(NOTE.-A C is said to join the extremities towards the same parts.

CB joins the extremities towards opposite parts.) Proof.

A B is parallel to CD (Given a).
B C 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 B C,

.'. by Axiom 2a, (e) the two sides A B, BC, are equal to the two sides DC, CB, 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 (f) the base AC equal to the base BD, and (8) the angle ACB equal to the angle D BC.

So, the straight line B C meets A C, BC,
and makes the alternate angles equal (8).
.. by Euc. I. 27,

AC is parallel to BC, and
AC is equal to BC (f). Q. E. D.

EXERCISE.
Given.-A B, CD parallel,

AC, B D equal.

B В

Required.—To prove that angles C AB, BDC

are together equal to two right angles. (Draw B E parallel to A C. Foin C B. Euc. I. 33, 1. 5, 1. 8, 1. 13, 1. 32.)

And use

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.)

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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.

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