5 to any number of decimal places desired. Thus, supposing AB=a=10,

And AH+HB=AB=6.18034+3.81966=10.00000.

COR. I. To cut a line in extreme and mean ratio, it must first be produced in extreme and mean ratio; that is, CF.FA must equal A B2.

COR. II. When a line CF, or its equal, is cut in extreme and mean ratio, the rectangle AC. (AC-AF) is equal to the square of AH, or AF; or AC. HB = AH2.

Hence, if a line, CF, be cut in extreme and mean ratio, the greater segment, AC, will be cut in the same manner, by taking in it a part equal to the less, AF; and the less, AF, will be similarly cut, by taking in a part equal to the difference (ACAF) or HB; and so on.

COR. III. A line CF being cut in extreme and mean ratio, the rect. AC.AF under its segments AC - AH', the difference between their squares; thus 10 x 6.18034 100-38.1966 =

61.8034.

=

SCH.-Let A be a line cut in extreme and mean ratio, G the greater segment, L the less, and D the difference: then 1. A2 +L2 = 3G2; 2. (A+L)2 =5G2; 3. A.D=G.L; and 4. L2=G.D.

USE AND APP.-This 11th Prop. is applied in 10. iv., to the drawing of an isosceles triangle, of which each of the angles at the base is double of the third angle; and, in 30. vi., to the cutting of a line in extreme and mean ratio,—that is, so that the whole line is to the greater segment as the greater is to the less. The construction of pentagons in bk. iv. also depends on this problem, and of regular bodies, as described in bks. xiii. xiv. and xv., called also the Platonic Solids.

PROP. 12.-THEOR.-(Important.)

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 st. line intercepted without the triangle between the perpendicular and the obtuse angle.

CONS.-12. I. To draw a st. line perpendicular to a given st. line of an unlimited length from a given point without it.

Pst. 2. A terminated st. line may be produced to any length in a st. line. DEM.-4. II. If a st. line be divided into any two parts, the square of the whole line is equal to the square of the two parts, together with twice the rectangle contained by the parts.

Ax. 2. If equals be added to equals, the wholes are equal.

47. I. In any rt. angled triangle, the square described upon the side subtending the rt. angle is equal to the square described upon the side containing the right angle.

Let ABC a triangle have

an obtuse ACB;

From A to BC produced
draw a perp. AD;
then AB (AC+ BC2)
by twice BC.CD.

For, since BD is divided
into two parts at C;
BC2+ CD2+

.. BD2 =

2 BC.CD:

BG C

A

adding AD' to each, BD2 + AD2 = BC2 +
CD2+AD2+2 BC.CD.

But ADB being a rt. ang., AB2 = BD2 +
AD2;

and also AC2 = CD2+ AD2;

therefore AB2= BC2 + AC2 + 2 BC. CD;
i. e., A B2 > A C2 + BC2 by twice BC.CD.
Wherefore, in obtuse angled triangles, if a
perpendicular, &c.

Q.E.D.

Alg. & Arith. Hyp.-Let BC=a=5; CA=b=6.708; AB=c=10; and

let CD=m=3; DA=n=6;

then BD=a+m=5+3=8.

the triangles ABD and ACD are rt. angled, viz., at ADB and ADC.

Alg. Since c2=(a+m)2+n2; and b2=m2 +n2 ;

therefore c2-b2 = (a + m)2 —m2 = a2 +2 am + m2 —m2 = a2 +2 am, and c2 = b2 + a2+2am; i. e., c2 is greater than b2 + a2 by twice a m.

Arith. Since 102 = (5+3)2 +62=64+36; & 6.7082 or 45=32+62=9+36; therefore 102-6.7082, or 100-45=(5+3)2-32-64-9=25+30;,

and 1026.7082 +52 + (2 × 5 × 3) = 45+25+30:

i. e., 102 or 100 greater than 45 +25 by 30.

USE AND APP.-1. By this Proposition the Area of a triangle may be ascertained when the three sides are known.

Let the sides A B=20, AC=13, and BC= 11.

Now (12. I.) AB2 = AC2 +BC2+2 BC.CD;

therefore AB2-AC2-BC2; or AB2-(AC2+B C2)=2 BC.CD; i. e., 2 BC.CD=400—(169+121)=400—290=110;

But (47. I.) AC2-CD2 = AD2; i. e., 169—25=144; and AD= √144

= 12.

We have now ascertained AD the altitude of the triangle ABC, and BC

the base is given;

Again (41. I.) the Area of a triangle equals half that of a parallelogram on the same base and of the same altitude:

Thus the Area of triangle ABC=

AD.BC 12 x 11
2
2

=

132
2

=66.

Or, by bisecting the line BC in G, we obtain the same result; thus,

2 BC 2

And DG+ =DB; then (47. I.) AB2-BD2 = AD2; whence we find

= 16; then (47. I.) 400-256=144; and √144 AD as

2

before.

PROP. 13.-Theor.

In every triangle, the square of the side subtending either of the acute angles is less than the squares of the sides containing that acute 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.

CONS.-12. I., Pst. 2.

DEM.-7. II. If a st. 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.

Ax. 2, 47. I.

16. I. If one side of a triangle be produced, the ext. angle is greater than either of the int. and opposite angles.

12. II. 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 con-
tained 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.

3. II. If a line be divided into any two parts, the rectangle contained
by the whole and one of the parts equals the rectangle contained
by the two parts together with the square of the aforesaid part.
Ax. 6. Doubles of the same are equal.

Ex.& 1 Hyp.

CONS. 2 12. I.

3 Concl.

Let A ABC have the B acute,

On BC one of the sides let the perp. AD fall from BAC;

then AC

(AB+BC) by twice BC. BD.

CASE I.-Let the perp. AD fall within the triangle ABC.

DEM. 1 C.2 & 7. II.

Since BC is divided at

D, CB+B D3 = C D2
+2 BC. BD;

2 Add. & Ax.2 On adding A D2, C B2+

BD2+ AD2 = CD2 +

AD2+2 BC.BD: B

3 C.2 & 47. I. But / sat D being rt. s

4 D. 2 & 3.
5 Concl.

=

D

AB=BD+AD, and A C CD' + AD"; .. C B2+ A B2 = AC2+2 BC. BD; that is, AC3<< (BC+AB) by twice BC. BD.

CASE II.-Let the perp. AD fall without the▲ ABC.

=AC+twice (BC+BC.CD).

4 C. & 3. II. But BD being divided in C, the BD. BC

=BC+BC.CD;

DEM. 5 Ax. 6.

6 Ax. 6.

7 D. 3 & 6.

8 Concl.

Now the doubles of equals being equal, twice BD. BC= twice BC. CD + twice BC2:

AB2+ BC2 = AC2 + twice BD. BC; |.. AC2 alone < AB2 + BC2 by 2 BD. BC.

CASE III-Lastly, let the side AC be perpendicular to BC.

DEM.1 47. I.

Here A B2 = AC2 + BC2;

2 Add. & Ax.2 and adding BC2, then A B2 + BC2 = AC2 +2 BC2, or 2 BC. BC.

3 Recap.

Wherefore, in any triangle, the square subtending, &c.

Q.E.D.

CASE I.-Alg. & Arith. Hyp. Let BC=a= 10; AB=c=√61=7.8102; AC =b=√/41=6.4031; BD=m=6; AD=n=5; and DC=a—m=10

-6=4.

Alg. (47. I.) c2 = n2 + m2 and b2 = n2 + (a—m)2

Subtracting, c2—b2 — m2 —(a—m)2 = m2 — (a2 —2 am+m)=2 am—a2
Transpose, a2+c2 = b2 +2 am; or b2+2am=a2 + c2

therefore, b2 is less than a2 + c2 by 2 am.

Arith. (47. I.) 61=25+36, and 41=25+16

Subtracting, 61-41 = (2 × 10 × 6)—100-120-100-20 Transposing, 100+61=41+120; or 41 +120=100+61 therefore, 41 is less than 100+61 by 120.

CASE II.—The perpendicular AD here passes out of the triangle, and the positions of C and D are changed; so that we have BC=a=2; AB =c=√61; AC=b= √41; BD=m=6; AD=n=5; and DC=m— a=6-2=4.

Alg. (47. I.) c2 = m2 +n2 and b2 = (m—a)2 +n2

Subtracting c2-b2 — m2 — (m—a)2 = m2 —m2 +2 am-a2 = 2am-a2
Transpose a2 + c2 = b2 +2 am; or b2+2am=a2+c2

therefore b2 is less than a2 + c2 by twice am

Arith. (47. I.) 61=36+25, and 41=16+25

Subtracting 61-41=36-16=36-36+24-4=(2× 12)—4

Transpose 4+61=41 + (2 × 12); or 41+24=4+61

therefore 41 is less than 4+61 by 24.

CASE III.-The perpendicular AD and the side AC coincide, and the points D and C; so that we have BC=a=4; AD=AC=b=5; and AB =c=√/41.

N