any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half line and the part produced.

Let the str. line AB be bisected in C, and produced to D; then AD. DB + CB2 =CD2.

.. gnomon CMG = AD. DB,

=

add to each, the fig. LG, which is equal to CB'. .. gnomon CMG + fig. LG AD.DB+CB2, i. e. whole fig. CEFD or CD2 = AD.DB+CB2. Wherefore, if a str. line, &c.

Note. If the half line be a, and the part produced b, the proposition algebraically expressed is (2a+b)b+a2=(a+b)2.

PROP. XLVI. THEOR. 7. 2 Eu.

If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the str. line AB be divided into any two parts at the pt. C; then AB2+BC2=2.AB.BC + AC2.

add to each the fig. CK.

fig. AK = fig. CE;

.. fig. AK+fig. CE 2 fig. AK

but, fig. AK + fig. CE=

.. gnomon AKF+ fig. CK

gnomon AKF +
fig. CK,
= 2. fig. AK,
2 fig. AK, for Cor.
Prop. 43.

but, 2 AB. BC ={2

BK = BC;

.. gnomon AKF + fig. CK = 2. AB. BC, add to each the fig. HF, which is equal to AC2, :. gnomon AKF + fig.

CK + fig. HF

but gnomon AKF + fig.

CK +fig. HF

=2. AB. BC+AC2,

.. whole fig. ADEB += 2 AB. BC+AC2,

i. e.

fig. CK

AB+BC2 2 AB. BC + AC2.

=

Wherefore, if a str. line, &c.

Note.-If the parts of the line be a and b, the proposition algebraically expressed is (a + b)2 + a2 = 2a (a + b)+b2. Õr (a + b)2 +b2=2b (a + b) + a2.

PROP. XLVII.

THEOR. 12. 2 Eu.

In obtuse angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle, is greater than the squares of 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 ABC have the obtuse ACB. then From A draw AD BC produced : BA2> BC + CA2 by 2 BC. CD.

. str. line BD is divided into 2 parts in pt. C, BD2 = BC+CD+2. BC. CD, Prop. 43.

but {

i. e.

add to each DA2

BD+DA=

[BC2 + CD2 + DA2 +
2 BC. CD;

BD2 + DA2 = BA3,

CD + DA2 = CA2,

BA2= BC+CA2+2 BC. CD,
BA2 > BC2+CA3 by 2 BC. CD.

Therefore, in obtuse angled As, &c.

Prop. 39.

PROP. XLVIII. THEOR. 13. 2 Eu.

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

Let ABC have the acute ABC, upon BC one of the sides containing it, draw AD 1 BC; then AC2 < CB2 + BA' by 2 CB.BD.

(Fig. 1.)

Prop. 46. ..

str. line CB is divided into 2 parts at D, CB2 + BD2 = 2. CB. BD + DC2 add AD2 to each

.. CB2+BD2+AD2 = 2.CB.BD+DC2+ AD3, BD2 + AD2 = BA2,

Prop. 39.

but

i. e.

{

DC2+ AD9

=

AC2,

CB+BA

2 CB. BD+AC2 AC2 < CB2+BA2 by 2 CB.BD. 2ndly. When AD falls without the ABC.

(Fig. 2.)

Prop. 15.

Prop. 47.

Prop. 42.

Ax. 6.

Prop. 39.

D= rt. ▲,:. ≤ACB > rt

AB AC+CB2+2 CB. CD, add CB2 to each,

.. AB2+CB2 AC2+2CB2+2CB. CD.
BD is divided into 2 parts at C.
BD.CB CB2+ CB. CD,

.. 2 BD. CB = 2 CB2+2. CB. CD,
.. AB2+CB2

i. e.

AC+2 BD. CB, AC2 < CB2 + AB2 by 2 BD. CB. Lastly. When ACI BC, then is BC the str. line between the perpendicular and the acute at B. (Fig. 3.)

And AB AC2 + CB2,

=

add CB2 to each.

.. AB2+CB2 = AC2+2 CB2,

i. e.

AC2 < AB2 + CB2 by 2 CB. CB.

Therefore in every ▲, &c.