Proof.

Angle F G H is equal to angle E (by construction). Angle K HM is equal to angle E (by construction). .. by Axiom 1,

angle F G H is equal to angle K H M. To each of these equals add the angle KH G. Then, by Axiom 2a,

angles FGH, GHK are equal to angles GHK, KH M.

But, by Euc. I. 29,

angles FGH, GHK are equal to two right angles. ... angles G H K, KH M are equal to two right angles.

Hence, by Euc. I. 14,

G H is in the same straight line with H M. Again, FK is parallel to GM, and

KH meets them.

... by Euc. I. 29,

angle FKH is equal to interior opposite angle KHM, To each of these add the angle HKL.

Then, by Axiom 2a,

angles M HK, H K L are equal to angles HKL, HK F.

But, by Euc. I. 29,

MHK, HKL are equal to two right angles. ... H KL, HK F are equal to two right angles.

Hence, by Euc. I. 14,

F K is in the same straight line with KL.

Again, FG is parallel to KH, and

LM is parallel to KH.
.. by Euc. I. 30,

F G is parallel to L M,
and FL is parallel to GM.
... by definition,
FGML is a parallelogram.

Again, the triangle A D B is equal to the parallelogram FH (by construction),

and the triangle DB C'is equal to the parallelogram KM (by construction). .. by Axiom 2,

The rectilineal figure A B C D is equal to the parallelogram FGML, which has been described with its angle FG M equal to the angle E. Q. E. F.

B

E

PROBLEM (Euclid I. 46).

Repeat. The enunciations of Euc. I. 29 and 34, Axiom 1, and definition of a square.

General Enunciation.

To describe a square on a given straight line.

Particular Enunciation.

Given. The straight A line A B.

Required. To describe a square on A B.

B

Construction.

(a) From the point A draw A Cat right angles to AB (by Euc. I. II).

(b) Make A D equal to AB (by
Euc. I. 3).

(c) Through D draw DE parallel
to AB (by Euc. I. 31).
(d) Through B draw BE parallel
to A B (by Euc. I. 31).
ADEB shall be a square.

୯

D

Proof.

By (c) and (d), A DEB is a parallelogram. by Euc. I. 34,

(e) A B is equal to D E, and AD to B E. But by (b),

Again, AB is parallel to D E, and

AD meets them.

A

-But BAD is a right angle.

.. ADE is also a right angle.
And by Euc. I. 34,

ABE, BED are right angles.

AB is equal to AD.
.. by Axiom 1,

(f) AB, BE, E D, and D A are all equal.

(g). angles BAD, ADE are equal to two right angles.

Hence, by (f) and (h),

The figure ABED is a square,

and it is described on the line A B.

(h) .. BAD, ADE, ABE, BED are all right angles.

Q. E. F.

(b) EB, BC are in the same straight line.

D

E B

Given. The right-angled

triangle ABC: of which (a) the angle BAC is the

right angle.

(c) AF is equal to E C.

(Use Euc. I. 15, then I. 14, to prove (a) and (b), and the definition of a square to prove (c).)

THEOREM (Euclid I. 47).

Repeat. The enunciations of Euc. I. 4, 14, 41; Axioms 2, 6a, and 11, and definition of a square. General Enunciation.

In any right-angled triangle, the square which is described on the side subtending the right angle, is equal to the squares on the sides which contain the right angle. Particular Enunciation.

Fig. 1

B

Required. To prove that
the square described on

BC shall be equal to the squares described on
BA, A C.

Construction.

On BC (by Euc. I. 46) describe the square

BCDE.

On BA (by Euc. I. 46)
ABFG.

describe the square

On AC (by Euc. I. 46)
АС К Н.

describe the square

(b) Through the point A, draw AL (Fig. 3) parallel to BD and CE (by Euc. I. 31). Join AD, FC.

Proof.

BAC is a right angle (a).

BAG is a right angle (definition).

by Euc. I. 14,

(c) CA is in the same straight line with AG.

BAC is a right angle (a).

CAH is a right angle (definition).