Let the straight line G H K cut A B, E F, CD. Then, because G H K cuts the parallel straight lines A B, E F, therefore the angle A G H is equal to the angle G H F (I. 29). Again, because G H K cuts the parallel straight lines E F, CD, therefore the angle G H F is equal to the angle H KD (I. 29). And it has been shown that the angle A G H is equal to the angle G H F,

therefore the angle A G H is equal to the angle G K D.
But these are alternate angles,

therefore A B is parallel to C D (I. 27).
Therefore, straight lines which are parallel, etc.

Proposition XXXI. Problem.

Q. E. D.

To draw a straight line through a given point parallel to a given straight line.

Let A be the given point, and B C the given straight line.
It is required to draw through A a straight line parallel to

At the point A, in the straight line A D,

make the angle D A E equal to the angle A D C (I. 23).

Produce E A to F.

Then EF shall be parallel to B C.

Because the straight line A D falls upon the two
straight lines E F, BC,

making the alternate angles E A D, A D C equal to one another, therefore E F is parallel to B C (I. 27).

Therefore, through the given point A, a straight line E A F has been drawn parallel to B C.

Proposition XXXII. Theorem.

Q. E. F.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles.

Let A B C be a triangle. Let one of its sides BC be produced to D.

Then the exterior angle A CD shall be equal to the two interior and opposite angles CA B, A B C, and the three interior angles A B C, B C A, C A B shall be equal to two right angles.

Through the point C draw C E parallel to B A (I. 31). Then, because CE is parallel to B A, and A C falls upon them, therefore the angle A CE is equal to the angle B A C (I. 29). Again, because CE is parallel to B A, and B D falls upon them, therefore the angle E C D is equal to the angle A B C (I. 29). But the angle A CE was proved equal to the angle B A C,

therefore the whole exterior angle A C D is equal to the two interior and opposite angles CA B, A B C (Ax. 2). To each of these equals add the angle A C B,

therefore the angles A CD, A C B are together equal to the

three angles CA B, ABC, ACB (Ax. 2).

But the angles AC D, ACB are equal to two right angles (I. 13),

therefore also the angles CA B, A B C, A C B are equal to two right angles (Ax. 1).

Therefore, if a side of any triangle be produced, etc.

COR. I. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

B

For any rectilineal figure A B C D E can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles.

Then, because the three interior angles of a triangle are equal to two right angles, and there are as many triangles as the figure has sides;

therefore all the angles of all these triangles are equal to twice as many right angles as the figure has sides.

But all the angles of all these triangles are equal to the interior angles of the figure, together with the angles at the point F.

And the angles at the point F are equal to four right angles (I. 15. Cor. 2).

Therefore all the angles of all these triangles are equal to the angles of the figure, together with four right angles.

But it has been proved that all the angles of all these triangles are equal to twice as many right angles as the figure has sides.

Therefore all the angles of the figure, together with four right

angles, are equal to twice as many right angles as the figure has sides.

COR. 2. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles.

A

For every interior angle A B C, with its adjacent exterior angle A B D, is equal to two right angles (I. 13).

Therefore all the interior angles, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure.

But, by the preceding corollary, all the interior angles, together with four right angles, are equal to twice as many right angles as the figure has sides.

Therefore all the interior angles, together with all the exterior angles of the figure, are equal to all the interior angles together with four right angles (Ax. 1).

From each of these equals take away all the interior angles, which are common to both.

Therefore all the exterior angles of the figure are equal to four right angles (Ax. 3).

Proposition XXXIII. Theorem.

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

Let A B, C D be equal and parallel straight lines, and joined towards the same parts by the straight lines A C, B D.

Then A C, B D shall be equal and parallel,

B

Join B C.

Then because A B is parallel to C D, and B C falls upon them therefore the angle A B C is equal to the angle B C D (I. 29). And, because, in the triangles A B C, D CB,

A B is equal to CD, and B C is common,

and the angle A B C is equal to the angle B CD; therefore the base A C is equal to the base B D (I. 4), and the triangle A B C to the triangle B C D,

and the remaining angles of the one to the remaining angles of the other.

Therefore the angle A C B is equal to the angle CBD. And, because the straight line B C falls upon the two straight lines AC, BD, and makes the alternate angles AC B, C B D equal to one another;

therefore A C is parallel to B D (I. 27);

and A C was shown to be equal to B D. Therefore, the straight lines which join the extremities, etc.

Q. E. D.

Proposition XXXIV. Theorem.

The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects it, that is, divides it into two equal parts.

Let A C D B be a parallelogram, of which B C is a diameter. Then the opposite sides and angles of the figure shall be equal to one another, and the diameter B C bisects it.