Pari. En. Let ABCbe a triangle right-angled at B; it is required to prove that AC is equal to AB2 + BC2.

Proof. On AB, BC, CA describe the squares ADEB, BFGC, CIHA respectively. Join CD, BH; and draw BJ parallel to AH. Since the angles ABC, ABE, BCF are right angles, it follows that CBE, ABF are straight lines; (1. 3.)

therefore the triangle DAC is on the same base DA, and between the same parallels DA, EC with the square DABE; therefore the triangle DAC is half the square DABE;

(II. 3, Cor. 2.) and similarly the triangle BAH is half the rectangle AJ.

But because the angle DAB = the angle HAC, each being a right angle; add to each the angle BAC;

therefore the whole angle DAC is equal to the whole angle BAH;

and the two sides DA, AC are respectively equal to the two sides BA, AH;

therefore the triangle DAC is equal to BAH;

(Constr.)

and therefore the square DABE = the rectangle AJ.

(1. 5.)

Similarly it may be shewn that the square BCGF= the rectangle CJ, and therefore, since AJ and CJ make up

the whole square AHIC, the square AHIC is equal to the sum of the squares ABDE and BCGF, that is,

AC2= AB2 + BC3.*

This important proposition may be proved as follows.

Place two squares EABH, KBCF, as in the figure, with their sides AB, BC continuous and in the same straight line. From CB cut off CD equal to AB. Join DE, DF. Produce BK to L, making KL = AB. Join LE, LF.

H

Then it will be easy to prove that the triangles EAD, DCF, EHL, LKF are all equal, being right-angled, and having the sides containing the right angle equal; therefore the figure LEDF is equal to the sum of the two given squares, and all its sides are equal.

And since EDA is complementary to AED or FDC, therefore the angle EDF is a right angle.

Therefore LEDF is a square, and is the square on ED.

Therefore the square on the hypotenuse ED is equal to the sum of the squares on the sides EA, AD,

*Euclid, I. 47.

COR. I. It follows that in a triangle ABC right-angled

at B,

AB2 = AC2 - BC and BC2 = AC2 – AB'.

The next two theorems shew the modifications which the theorem undergoes when the triangle is not right-angled.

THEOREM IO.

In an obtuse-angled triangle the square on the side subtending the obtuse angle is greater than the squares on the sides containing that angle by twice the rectangle contained by either of these sides and the projection on it of the other side*.

Part. En. Let ABC be the triangle, ABC being the obtuse angle, BD the projection of AB on BC, BC being produced backward.

Then will AC2 = AB2 + BC2 + 2 CB. BD,

for

but

and

therefore

AC2 = AD2 + DC2, by (II. 9,)

AD2 = AB2 - BD3,

DC2=CB+BD + 2CB. BD, (by II. 6,)

AC2 = AB2 + BC2 + 2 CB.BD.

*Euclid, 11. 12.

THEOREM II.

In any triangle the square on the side opposite an acute angle is less than the squares on the other two sides by twice the rectangle contained by either side and the projection on it of the other side*.

Part. En. Let ABC be a triangle, B an acute angle, BD the projection of AB on BC, then will

AC2 = AB2 + BC2 - 2 CB × BD.

A

but

and

therefore

COR.

AD-AB-BD, by the same Theorem,
DC2 = BC2 + BD2 – 2CB × BD, (by 11. 7);

AC2 = AB2 + BC2 - 2 CB × BD.

Conversely, the angle opposite a side of a triangle is an acute angle, a right angle, or an obtuse angle, according as the square on that side is less than, equal to, or greater than the sum of the squares on the other two sides.

THEOREM 12.

The sum of the squares on two sides of a triangle is double the sum of the squares on half the base and on the line joining the vertex to the middle point of the base.

* Euclid, II. 13.

Part. En. Let AC, a side of the triangle ABC, be bisected in D; then

will

AB2 + BC= 2AD2 + 2BD3.

Proof. For let DE be the projection of BD on AC.

Then AB2 = AD2 + DB2 + 2AD. DE (by II. 10), and BC2 = CD2 + DB2 – 2CD. ED (by II. 11), therefore remembering that AD = DC, we obtain by addition that

AB2 + BC2 = 2AD2 + 2DB2.

This theorem in a more general form is known as Apollonius's Theorem.

THEOREM 13.

If a straight line is divided internally or externally at any point, the sum of the squares on the segments is double the sum of the squares on half the line and on the line between the point of division and the middle point of the line*.

Let AB be bisected in C, and divided internally or externally in D.

A

D

B

A

B

Then the squares on AD, DB will be double of the squares on, AC, CD.

Proof. For AD2 = AC+ CD2 + 2AC × CD by 11. 6;
and DB2 = CB2 + CD3 − 2BC × CD by II. 7;

therefore, adding, and remembering that AC= BC, and that therefore the rectangle ACx CD= the rectangle BC x CD, we get that AD3 + DB2 = 2AC + 2 CD3,

* Eucl. II. 9, 10.