sected, and of the square on the line made up of the half and the part produced.

D.

Let the straight line A B be bisected in C and produced to

Then the squares on A D, D B shall be double of the squares on A C, C D.

From the point C draw CE at right angles to A B (I. 11); make C E equal to A C or C B (I. 3) ; join A E, AB.

Through E draw E F parallel to A B (I. 31)

and through D draw D F parallel to C E.

Then, because E C is parallel to FD and E F falls upon them, therefore the angles CE F, E F D are equal to two right angles, and therefore the angles B E F, EFD are less than two right angles.

But if a straight line meets two straight lines so as to make the two interior angles on the same side together less than two right angles, these straight lines, being continually produced, shall at length meet upon that side on which are the

angles which are less than two right angles (Ax. 12) ;
therefore E B, F D will meet if produced towards B, D.
Let these be produced and meet in G ; join A G.
Then, because CA is equal to CE,

therefore the angle CA E is equal to the angle CEA (I. 5). But the angle ACE is a right angle,

therefore each of the angles CA E, CE A is half a right angle. For the like reason each of the angles C B E, CE B is half a right angle;

therefore A E B is a right angle.

And because the angle D B G is half a right angle,

for it is equal to the angle E BC (I. 15),
and that B D G is a right angle,

for it is equal to the alternate angle D CE (I. 29), therefore the remaining angle D G B is half a right angle ; therefore the angle D G B is equal to the angle D B G, and therefore the side D G is equal to the side B D (I. 6). Again, because F G E is half a right angle, and E F G is a right angle,

for it is equal to the opposite angle E CD (I. 34), therefore the remaining angle F E G is half a right angle; therefore the angle F E G is equal to the angle F G E, and therefore the side F E is equal to the side F G (I. 6). And because A C is equal to CE,

therefore the square on A C is equal to the square on CE, and therefore the squares on A C, CE are double of the square on A C.

But the square on A E is equal to the squares on A C, CE, therefore the square on A E is double of the square on A C. Again, because E F is equal to F G,

therefore the square on E F is equal to the square on F G, and therefore the squares on E F, F G are double of the square on E F.

But the square on E G is equal to the squares on EF, FG; therefore the square on E G is double of the square on E F ; and E F is equal to C D (I. 34)

therefore the square on E G is double of the square on C D. And it has been demonstrated

that the square on A E is double of the square on A C, therefore the squares on A E, E G are double of the squares on AC, CD.

But the square on AG is equal to the squares on A E, EG (I. 47),

therefore the square on A G is double of the squares on A C, C D. And the square on A G is equal to the squares on A D, D G, therefore the squares on A D, D G are double of the squares on A C, CD.

And D G is equal to D B,

therefore the squares on A D, D B are double of the squares on

AC, CD.

Therefore, if a straight line, etc.

Proposition XI. Problem.

Q. E. D.

To divide a straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square on the other part.

Let A B be the given straight line.

It is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square on the other part.

Upon A B describe the square A C D B (I. 46)

bisect A C in E (I. 10); join B E.

Produce C A to F, making E F equal to E B (I. 3). Upon A F describe the square A F G H (I. 46). Then A B shall be divided in H so that the rectangle A B, BH is equal to the square on A H.

Produce G H to meet C D in K.

Then, because A C is bisected in E and produced to F, therefore the rectangle CF, F A, together with the square on

A E,

is equal to the square on E F (II. 6).

But E F is equal to E B ;

therefore the rectangle C F, FA, together with the square on

A E,

is equal to the square on E B.

And the square on E B is equal to the squares on A E, A B

(I. 47),

because E A B is a right angle ;

therefore the rectangle CF, FA, together with the square on

A E,

is equal to the squares on A E, A B.
Take away the common square on A E;
therefore the remainder, the rectangle C F, FA,
is equal to the square on A B.
But the figure F K is the rectangle C F, FA,
for F G is equal to FA;
and AD is the square on A B;
therefore F K is equal to A D.

Take away the common part A K,

therefore the remainder F H is equal to the remainder H D. But HD is the rectangle A B, B H,

for A B is equal to B D ;

and F H is the square on A H;

therefore the rectangle A B, B H is equal to the square on A H. Therefore the given straight line A B is divided in H so that the rectangle A B, B H is equal to the square on A H.

Proposition XII. Theorem.

Q. E. F.

In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle.

Let A B C be an obtuse-angled triangle, having the obtuse

angle A C B, and from the point A let AD be drawn perpendicular to B C produced.

Then the square on A B shall be greater than the squares on A C, CB by twice the rectangle B C, CD.

Because the straight line B D is divided into two parts in C, therefore the square on B D is equal to the squares on B C, C D and twice the rectangle B C, C D (II. 4).

To each of these equals add the square on D A,

therefore the squares on B D, DA are equal to the squares on B C, CD, DA,

and twice the rectangle B C, C D.

But the square on B A is equal to the squares on B D, D A (I. 47), because the angle at D is a right angle;

and the square on CA is equal to the squares on CD, DA. Therefore the square on B A is equal to the squares on B C, CA, and twice the rectangle B C, CD;

that is, the square on B A is greater than the squares on BC, CA, by twice the rectangle B C, C D.

Therefore, in obtuse-angled triangles, etc.

Proposition XIII. Theorem.

Q. E. D.

In every triangle, the square on the side subtending either of the acute angles is less than the squares on the sides containing that angle by twice the rectangle contained by either of these sides, and the straight line intercepted between the acute angle and the perpendicular let fall upon it from the opposite angle.

Let ABC be any triangle, and the angle at B one of its