But the last three triangles BEC, AEC, AEB, are together, equal to the whole triangle ABC; whence

x2/3=ax+bx+cx.

And, consequently, if each side of this equation be divided by x, we shall have

Which is, therefore, half the length of either of the three equal sides of the triangle.

COR. Since, from what is above shown, ad is = x√3, it follows, that the sum of all the perpendiculars, drawn from any point in an equilateral triangle to each of its sides, is equal to the whole perpendicular of the triangle.

PROBLEM VI.

Through a given point P, in a given circle ACBD, to draw a cord CD, of a given length.

Draw the diameter APB; and put CD α, AP = b', FB-c, and cp=x; then will p»=ɑ—x.

But, by the property of the circle, (Euc. 11. 35,) CP XPD AP XPB; whence

x(a-x)=bc, or
x2 -ɑœ――bc.

Which equation, being resolved in the usual way

gives

x={a±√(‡a3—bc) ;

Where has two values, both of which are positive.

PROBLEM VII.

Through a given point P, without a given circle ABDC, to draw a right line so that the part CD, intercepted by the circumference, shall be of a given length.

Draw PAB through the centre o; and put CD = a, PA=b, PB c, and rcx; then will PD=x+a.

But, by the property of the circle, (Euc. 111, 36, cor.,) PC XPD PA XPB; whence

x(x+a)=bc, or
x2+ax=bc..

Which equation being resolved, as in the former problem, gives

Where one value of x is positive and the other negative (c).

(c) The two last problems, with a few slight alterations, may be readily employed for finding the roots of quadratic equations by construction; but this, as well as the methods frequently given

Y

PROBLEM VIII.

The base BC, of any plane triangle ABC, the sum of the sides AB, AC, and the line AD, drawn from the vertex to the middle of the base, being given, to determine the triangle.

D

Put BD or DC=α, AD=b, AB+AC=8, and AB; then will AC-S-x.

But, by my Geometry, B. 11, 19, AB2+AC 2BD + 2AD3; whence

Which last equation, being resolved as in the former instances, gives

x={s±√(a2+b2 - 482),

for the values of the two sides AB and AC of the triangle ; taking the sign + for one of them, and for the other and observing that a3+b2 must be greater than s2.

PROBLEM IX.

The two sides AB, AC, and the line AD, bisecting the vertical angle, of any plane triangle, ABC, being given to find the base BC.

for constructing cubic and some of the higher orders of equa tions, is a matter of little importance in the present state of mathematical science; analysis, in these cases, being generally thought a more commodious instrument than geometry.

Put AB=α, Ac=b, AD=c, and вс=x; then, by Euc. vi, 3, we shall have

AB(α) : ac(b) :: BD : DC.

And, consequently, by the composition of ratios (Euc. v, 18,)

But, by Euc. vI, 13, BD XDC+ADAB XAC; wherefore,

Which is the value of the base BC, as required.

PROBLEM X.

drawn

Having given the lengths of two lines AD, BE, from the acute angles of a right angled triangle ABC, to the middle of the opposite sides, it is required to determine the triangle.

A.

E

D

Put AD α, BE = =b, сD or cв=x, and CE or ca=y; then, since (Euc. 1, 47) CD+CA2=AD3, and CE2+CB2 BE2, we shall have

Whence, taking the second of these equations from four times the first, there will arise

And, in like manner, taking the first of the same equations from four times the second, there will arise

Which values of x and y are half the lengths of the base and perpendicular of the triangle; observing that b must be less than 2a, and greater than α.

PROBLEM. XI.

Having given the ratio of the two sides of a plane triangle ABC, and the segments of the base, made by a perpendicular falling from the vertical angle, to determine the triangle.