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Since
A B= 2CB,

Cons. and

ED= 2 C B,
(the diameter of a O being twice the radius),
AB= ED.

Ax. 1
.. A D A B = AD - ED= A E.
But
A E= A H,

Cons. . AD AB= A H.

Ax. 1
Also AB - A E = AB A H= H B.
Substitute these equivalents in the last proportion.
Then AH : AB :: HB : AH.
Whence, by inversion, A B : AH :: AH : HB. $ 263
.. A B is divided at H in extreme and mean ratio.

Q. E. F.

REMARK. A B is said to be divided at H, internally, in extreme and mean ratio. If B A be produced to H', making A H' equal to A D, A B is said to be divided at H', externally, in extreme and mean ratio.

Prove AB: A H' :: AH' : H' B.

When a line is divided internally and externally in the same ratio, it is said to be divided harmonically.

Thus AB A_ Z__B__ Dis divided harmonically at C and D, if C A :CB::DA :D B; that is, if the ratio of the distances of C from A and B is equal to the ratio of the distances of D from A and B.

This proportion taken by alternation gives :

AC:AD::BC:BD; that is, C D is divided harmonically at the points B and A. The four points A, B, C, D, are called harmonic points; and the two pairs A, B, and C, D, are called conjugate points.

Ex. 1. To divide a given line harmonically in a given ratio.

2. To find the locus of all the points whose distances from two given points are in a given ratio.

PROPOSITION XXVIII. PROBLEM. 312. Upon a given line homologous to a given side of a given polygon, to construct a polygon similar to the given polygon.

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ВІ

C' Let A' E' be the given line, homologous to A E of the

given polygon A B C D E.

It is required to construct on A' E' a polygon similar to the given polygon.

Frone E draw the diagonals E B and EC.

From E' draw E' B', making 2 A' E' B' = A E B. Also from Al draw A' B', making 2 B' A' E' =_ BA E,

and meeting E' B' at B'. The two A A B E and A' B' E' are similar, $ 280 (two A are similar if they have two of the one equal respectively to two É

of the other). Also from E' draw E' C', making 2 B' E' C' = LBEC. From B' draw B'C', making _ E' B'C' = Z EBC,

and meeting E' C' at C'. Then the two A EBC and E' B' CTM are similar, § 280 (two A are similar if they have two ts of the one equal respectively to two

of the other). In like manner construct A E C'D' similar to AEC D.

Then the two polygons are similar, § 293 (two polygons composed of the same number of A similar to each other and

similarly placed, are similar).
.. A' B' C' D' E' is the required polygon.

Q. E,

F.

EXERCISES. 1. A B C is a triangle inscribed in a circle, and B D is drawn to meet the tangent to the circle at A in D, at an angle A BD equal to the angle A BC; show that A C is a fourth proportional to the lines B D, A D, A B.

2. Show that either of the sides of an isosceles triangle is a mean proportional between the base and the half of the segment of the base, produced if necessary, which is cut off by a straight line drawn from the vertex at right angles to the equal side.

3. A B is the diameter of a circle, D any point in the circumference, and C the middle point of the arc A D. If A C, A D, BC be joined and A D cut B C in E, show that the circle circumscribed about the triangle A E B will touch A C and its diameter will be a third proportional to B C and A B.

4. From the obtuse angle of a triangle draw a line to the base, which shall be a mean proportional between the segments into which it divides the base.

5. Find the point in the base produced of a right triangle, from which the line drawn to the angle opposite to the base shall have the same ratio to the base produced which the perpendicular has to the base itself.

6. A line touching two circles cuts another line joining their centres; show that the segments of the latter will be to each other as the diameters of the circles.

7. Required the locus of the middle points of all the chords of a circle which pass through a fixed point.

8. O is a fixed point from which any straight line is drawn meeting a fixed straight line at P; in 0 P a point Q is taken such that O Q is to 0 P in a fixed ratio. Determine the locus of Q.

9. O is a fixed point from which any straight line is drawn meeting the circumference of a fixed circle at P; in 0 P a point Q is taken such that 0 Q is to 0 P in a fixed ratio. Determine the locus of Q.

BOOK IV.

COMPARISON AND MEASUREMENT OF THE SUR

FACES OF POLYGONS.

PROPOSITION I. THEOREM. 313. Two rectangles having equal altitudes are to cach other as their bases.

C D

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.

Let the two rectangles be AC and A F, having the

the same altitude A D.
We are to prove

rect. AC AB

rect. A F A E'
CASE I. When A B and A E are commensurable.
Find a common divisor of the bases A B and A E, as A 0.

Suppose A O to be contained in A B seven times and in A E four times.

A B 7
Then

A E =Ā
At the several points of division on A B and A E erect Is.

The rect. AC will be divided into seven rectangles,

and rect. A F will be divided into four rectangles.

These rectangles are all equal, for they may be applied to each other and will coincide throughout.

.: rect AC 7

rect AF But

A È

A B

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CASE II. — When A B and A E are incommensurable.

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Divide A B into any number of equal parts, and apply one of these parts to A E as often as it will be contained in A E.

Since A B and A E are incommensurable, a certain number of these parts will extend from A to a point K, leaving a remainder K E less than one of these parts.

Draw K H il to E F.
Since A B and A K are commensurable,
rect. AH A K

Case 1 rect. AC ĀB' Suppose the number of parts into which A B is divided to be continually increased, the length of each part will become less and less, and the point K will approach nearer and nearer to E.

The limit of A K will be A E, and the limit of rect. A H will be rect. A F.

.: the limit of 4 K will be 4 E,

rect. A Ċ will be rect. A F

and the limit of rect. A

rect. A C
40" "co
Now the variables A K and rect. A H.
and A Banu reet. A C.

are always equal however near they approach their limits; .. their limits are equal, namely,

rect. AF AE

$ 199 rect. A C - A B' S

Q. E. D. 314. COROLLARY. Two rectangles having equal bases are to each other as their altitudes. By considering the bases of these two rectangles A D and A D, the altitudes will be A B and A E. But we have just shown that these two rectangles are to each other as A B is to A E. Hence two rectangles, with the same base, or equal bases, are to each other as their altitudes.

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