Proof.

ABCD is a parallelogram.
.. by Euc. I. 34,

(a) AD is equal to BC.

EBCF is a parallelogram.
by Euc. I. 34,

(b) EF is equal to B C.

Hence, by (a) and (b), and Axiom 1, (c) A D is equal to É F.

To each of these equals add DE.
Then, by Axiom 2a,

(d) the whole AE is equal to the whole DF. ABCD is a parallelogram.

.. by Euc. I. 34,

(e) A B is equal to DC.

Hence, by (d) and (e),

(f) the two sides E A, A B are equal to the two sides F D, DC, each to each.

Again, AB is parallel to DC.

.. by Euc. I. 29,

(g) angle FDC is equal to the interior opposite angle E A B.

Hence, by (f) and (g), and Euc. I. 4,

(h) The triangle EAB is equal to the triangle FD C.

ABCF is a trapezium,

take the equal triangles () from the trapezium ABCF,

Then, by Axiom 3,

Q. E. D.

The remainder ABCD is equal to the remainder EBC F. EXERCISE.-General Enunciation (PART III.). Given. The parallelograms ABCD, EBCF,

on the same base BC, and between the same parallels A and B C. But having the side EF partly upon the side AD (see next page).

Required. To prove that A B CD is equal to

A

B

EBCF (Adopt the same method as in Case II. of preceding Theorem.)

THEOREM (Euclid I. 36).

Repeat. The enunciations of Euc. I. 33, 34, 35, and Axiom I.

General Enunciation.

Parallelograms on equal bases and between the same parallels are equal to one another.

Particular Enunciation.

Given. The two parallelograms, ABCD, EFGH,

(a) on the equal bases BC, and FG, and between the same parallels A H and B G. Required. To prove that A B CD is equal to EFGH

Construction.

Join BE and CH.

Again, EH and BĊ are parallel, and they are joined towards the same parts by BE, CH. by Euc. I. 33,

(d) BE and C H are equal and parallel, and EH and B C are equal (c) and parallel.

.. by definition,

(e) EBCH is a parallelogram.

Again, A B C D and E B CH are on the same base and between the same parallels.

by Euc. I. 35,

(f) ABCD and E BCH are equal.

And EFGH and EB CH are on the same base

E B, and between the same parallels.

.. by Euc. I. 35,

(g) É F G H and E BCH are equal.

Hence, by (f) and (g), and Axiom 1,
ABCD and E F G H are equal.

Q. E. D.

I. EXERCISE (Euc. I. 37).

Given. The triangles A B C, D B C on the same base B C, and between the same parallels AD and B C.

A

B

Required. To prove that A B C and D B C are equal. Complete the parallelograms by producing AD both ways, and drawing lines parallel to A B, and DC, and use Euc. I. 34 and 35 for the proof.)

II. EXERCISE (Euc. I. 38).

Given. The triangles A B C and DEF, on the equal bases B C and E F, and between the same parallels AD and BF.

Required. To prove that A B C and D E F are equal. (Complete the parallelograms as in the preceding Exercise, and use Euc. I. 34 and 36.)

THEOREM (Euclid I. 39).

Repeat. The enunciations of Euc. I. 31, I. 37, and Axiom 1.

General Enunciation.

Equal triangles on the same base and on the same side of it are between the same parallels.

Particular Enunciation.

(b) Let us suppose that A E is parallel to BC.

Then AB C, E CB are two triangles on the same base B C, and between the same parallels A E and B C.

by Euc. I. 37,

(c) triangle ABC is equal to triangle ECB.