PROP. XI.

118. Demonstrate this by Algebra.

Let AB = a, AH = x, then HB = a

X,

The above expression for x has two values, : √5 is positive or negative; what results from its having two values?

That there are two solutions of the question when considered generally, that is, analytically. The first solution, in supposing 5 positive, corresponds to Euclid's construction. The second, in making √5 negative, gives the point H in BA produced, at a distance from A equal to

Instead of making use of Prop. XII. Book II. as Euclid does, the second case may be proved by Prop. VII., in the same words exactly as the first case, with the exception of the commencement, viz.

the | BD (Euclid's second fig.) is divided into two parts in the point D; after which

".. (7, Book II.) CB2 + BD2 = 2CB × BD + CD2, to each add AD2.

:. CB2 + BD2 + AD2 = 2CB × BD + CD2 + AD2.

But

AB2 = BD2 + AD2: L D is a L;

and AC2 CD2 + AD2,

:. CB2 + AB2 = AC2 + 2CB × BD,&c.”
which is Euclid verbatim.

ВООК III.

120. What is the subject of this Book? It treats of the properties of Circles.

DEFINITION IV.

121. What is the "distance" between two points?

In language,

It is the length of a line, whose extremities are these points.

But in Geometry,

It is the length of the straight line, whose extremities are these points.

122. What is the shortest line that can be drawn from one point to another?

Let A, B be the points, AB a straight line, and ACB a curved line joining them. AB is shorter than A CB.

D

B

In ACB take any point C, join AC, BC. Then (XX. B. I.)
AC+ CB > AB.

Again in AC take any point D, and in BC take any point E, &c. Then

AD + DC > AC and CE + EB > CB.

:. AD + DC + CE + EB > AC + BC and much > AB.

Hence proceeding we show that every exterior line ADCEB is than the next interior viz. ACB, and consequently it is evident that the curve is > AB.

In the above figure the curve is wholly concave towards

AB, but if it be partly convex and partly concave; partly rectilinear as well as curvilinear; partly above and partly below the line AB, the same method will apply, with very slight modifications.

From which then it follows that the shortest line that can be drawn from one point to another is the straight line. The above is the method of Legendre.

123. What is the shortest line that can be drawn from a given point to a given |?

Let A be the given point.

BC the given |. From A let

fall a AD upon BC, take B G FE

H C

any points E, F, G, H, and join on either side of D, AE, AF, AG, AH, &c.

Then because ADE is a L, AED is less than a L, and the greater is subtended by the greater side,

Again, since

.. AE is > AD.

LAEF is > ADE > L,

. AFE is < than a L.

:: ▲ AEF is > the AFE.

..(XIX., Book I.), AF is > AE, and much > AD.

Hence by similarly proceeding it follows that of all the straight lines which can be drawn from the point A to the | BC, the AD is the least, and that which is nearer to the is always less than that which is more remote.

124. Which of the points of a given is nearest to a given point?

Let A be the given point, and BC the given |. Then from the preceding article it appears that D is the point required.

Hence is suggested the following definition,

125. How is defined the distance of a given | from a given point?

It is the distance of the nearest point of the given | from the given point.

These preliminaries being explained, it becomes easy to perceive why Euclid has defined straight lines which are equally distant from the centre of a to be those to which L' drawn to them from the centre are =

DEFINITION XI.

126. What is the ambiguity of this definition?

The definition involves Prop. XXI. Book III., for it supposes that all the of either segment are to one another.

PROP. I.

127. Does Euclid's demonstration apply to all positions of the point G?

When G is taken in the | CE it does not, because then the FDB is not > GDB.

This case, however, is almost self-evident, because if G, the supposed centre,be any point in CE other than F, GC would be GE; but FC FE, from which an absurdity immediately follows.

=

If G be in CE it is evidently the middle point of CE; that is, it is F.

PROP. I. COR.

128. Hence it is also clear that if two straight lines, such as CE, be drawn, the centre of the circle, being in each of them, must be their point of intersection; which gives a practical method of finding the centre of a .