Because the angle E is equal to each of the angles FKH, GHM, therefore the angle FKH is equal to the angle GHM;

add to each of these the angle KHG;

therefore the angles FKG, KHG, are equal to the angles KHG, GHM; but FKH, KHG are equal to two right angles; (I. 29)

therefore also KHG, GHM, are equal to two right angles;

and because at the point H in the straight line GH, the two straight lines KH, HM, upon the opposite sides of it, make the adjacent angles equal to two right angles,

therefore KH is in the same straight line with HM. (I. 14.)

And because the straight line HG meets the parallels KM, FG, therefore the alternate angles MHG, HGF, are equal; (1. 29) add to each of these the angle HGL;

therefore the angles MHG, HGL, are equal to the angles HGF, HGL; but the angles MHG, HGL, are equal to two tight angles; (1. 29) wherefore also the angles HGF, HGL, are equal to two right angles, and therefore FG is in the same straight line with GL. (1. 14.)

And because KF is parallel to HG, and HG to ML,

therefore KF is parallel to ML; (1. 30)

and KM, FL, are parallels;

wherefore KFLM is a parallelogram; (1. 34, def.)

and because the triangle ABC is equal to the parallelogram HF, and the triangle ADC to the parallelogram GM;

therefore the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM.

Wherefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the angle E. Q. E. F.

Cor. From this it is manifest how to apply to a given right line a parallelogram, which shall have an angle equal to a given rectilineal

angle, and shall be equal to a given rectilineal figure-viz., by applying to the given straight line a parallelogram equal to the first triangle ABC, and having an angle equal to the given angle.

PROP. XLVI.-PROBLEM.

To describe a square upon a given straight line.

(References-Prop. 1. 3, 11, 29, 31, 34; ax. 1, 3; def. 30.)

Let AB be the given straight line.

It is required to describe a square upon AB.

From the point A draw AC at right angles to AB, (1. 11)

and make AD equal to AB, (1. 3)

through the point D draw DE parallel to AB, (1. 31)

and through B draw BE parallel to AD.

Then ADEB shall be the square required.

DEMONSTRATION

Because DE is parallel to AB, and BE parallel to AD; (constr.) therefore ADEB is a parallelogram; (1. 34, def.)

wherefore AB is equal to DE, and AD to BE; (1. 34)

but BA is equal to AD; (constr.)

therefore the four straight lines BA, AD, DE, EB, are equal to one. another,

and the parallelogram ADEB is equilateral.

Likewise all its angles are right angles;

for, since the straight line AD meets the parallels AB, DE,

therefore the angles BAD, ADE are equal to two right angles; (1. 29) but BAD is a right angle; (constr.)

therefore also ADE is a right angle; (ax. 3)

but the opposite angles of parallelograms are equal;' (1. 34) therefore each of the opposite angles ABE, BED, is a right angle;

wherefore the parallelogram ADEB is rectangular. And it has been demonstrated that it is equilateral; therefore ADEB is a square,

and it is described upon the given straight line AB.

Q. E. F.

Cor. Hence every parallelogram that has one right angle has all its angles right angles.

PROP. XLVII.-THEOREM.

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

(References-Prop. I. 4, 14, 31, 41, 46; ax. 2, 6, 11; def. 30.) Let ABC be a right-angled triangle, having the right angle BAC. Then the square described upon the side BC, shall be equal to the squares described upon BA, AC.

On BC describe the square BDEC, (1. 46)

and on BA, AC, the squares GB, HC;

through A draw AL parallel to BD, or CE, (1. 31) and join AD, FC.

DEMONSTRATION

Then, because each of the angles BAC, BAG, is a right angle, (hyp., def. 30)

the two straight lines AC, AG, upon the opposite sides of AB, make with it at the point A, the adjacent angles equal to two right angles; therefore CA is in the same straight line with AG; (I. 14)

for the same reason,

AB and AH are in the same straight line.

And because the angle DBC is equal to the angle FBA, each of them

being a right angle,

add to each the angle ABC,

then the whole angle DBA is equal to the whole FBC. (ax. 2)

Hence, in the two triangles ABD, FBC,

because the two sides AB, BD, are equal to the two FB, BC, each to each, (def. 30)

and the angle DBA equal to the angle FBC;

therefore the base AD is equal to the base FC,

and the triangle ABD to the triangle FBC. (1.4)

Again, because the parallelogram BL and the triangle ABD are upon the same base BD, and between the same parallels BD, AL; therefore the parallelogram BL is double of the triangle ABD; (1. 41) and because the square GB and the triangle FBC are upon the same base FB, and between the same parallels FB, GC;

therefore the square GB is double of the triangle FBC; but the doubles of equals are equal to one another; (ax. 6)

therefore the parallelogram BL is equal to the square GB.

In the same manner, by joining AE, BK, it is proved that the parallelogram CL is equal to the square HC.

Therefore the whole square BDEC is equal to the two squares GB, HC : (ax. 2.)

and the square BDEC is described upon the straight line BC,

and the squares GB, HC, upon BA, AC;

wherefore the square upon the side BC, is equal to the squares upon the sides BA, AC.

Therefore, in any right-angled triangle, &c.

Q. E. D.

PROP. XLVIII.-THEOREM.

If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it;

then the angle contained by these two sides is a right angle.

(References -Prop. 1. 3, 8, 11, 47; ax. 1, 2.)

Let the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC.

Then the angle BAC shall be a right angle.

D

B

CONSTRUCTION

From the point A draw AD at right angles to AC, (1. 11) and make AD equal to BA, (1. 3) and join DC.

DEMONSTRATION

Then, because DA is equal to AB,

the square of DA is equal to the square of AB;

to each of these equals add the square of AC,

therefore the squares of DA, AC, are equal to the squares of BA, AC; (ax. 2)

but because DAC is a right angle, (constr.)

herefore the square of DC is equal to the squares of DA, AC; (1. 47) and the square of BC is equal to the squares of BA, AC; (hyp.) therefore the square of DC is equal to the square of BC; (ax. 1) wherefore also the side DC is equal to the side BC.

Hence, in the two triangles DAC, BAC,

because the side DA is equal to the side AB, (constr.)

and AC common to both triangles,

the two DA, AC, are equal to the two BA, AC, each to each, and the base DC has been proved to be equal to the base BC;

therefore the angle DAC is equal to the angle BAC; (1. 8) but DAC is a right angle; (constr.)

therefore also BAC is a right angle. (ax. 1.) Therefore, if the square, &c.

Q. E. D.