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PROPOSITION XXV. PROBLEM. 306. To find a mean proportional between two given lines.

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Let the two given lines be m and n.
It is required to find a mean proportional between m and n.
On the straight line A E

take A C = m, and C B = n.
On A B as a diameter describe a semi-circumference.

At C erect the IC H.
CH is a mean proportional between m and n.

Draw H B and H A.
The 2 A H B is a rt. Z,

§ 204 (being inscribed in a semicircle), and HC is a I let fall from the vertex of a rt. Z to the hypotenuse.

..AC : CH :: CH : C B, $ 289 (the I let fall from the vertex of the rt. Z to the hypotenuse is a mean pro

portional between the segments of the hypotenuse).
Substitute for AC and C B their equals m and n.
Then
m : C H :: C H : .

Q. E. F.

307. COROLLARY. If from a point in the circumference a perpendicular be drawn to the diameter, and chords from the point to the extremities of the diameter, the perpendicular is a mean proportional between the segments of the diameter, and each chord is a mean proportional between its adjacent segment and the diameter.

PROPOSITION XXVI. PROBLEM. 308. To divide one side of a triangle into two parts proportional to the other two sides.

B E

Let A B C be the triangle. It is required to divide the side B C into two such parts that the ratio of these two parts shall equal the ratio of the other two sides, A C and A B. Produce C A to F, making A F= A B.

Draw FB.
From A draw A E Il to FB.

E is the division point required.
For
СА СЕ

§ 275 A FEB (a line drawn through two sides of a A Il to the third side divides those sides

proportionally).
Substitute for A F its equal A B.

CA CE
A B = EB

Q. E. F.
309. COROLLARY. The line A E bisects the angle C A B.
For
ZF=LA BF,

§ 112
(being opposite equal sides).
ZF= _ CA E,

$ 70
(being ext.-int. É).
ZABF= Z BAE,

(being alt.-int. E).
..Z CA E= Z B A E.

Ax. 1

Then

§ 68

310. DEF. A straight line is said to be divided in extreme and mean ratio, when the whole line is to the greater segment as the greater segment is to the less.

PROPOSITION XXVII. PROBLEM.

311. To divide a given line in extreme and mean ratio.

Ан в

Let A B be the given line.
It is required to divide A B in extreme and mean ratio.

At B erect a I BC, equal to one-half of A B.

From C as a centre, with a radius equal to C B, describe a O.

Since A B is I to the radius C B at its extremity, it is tangent to the circle.

Through C draw A D, meeting the circumference in E and D.

On A B take A H = A E.

H is the division point of A B required.

For AD : AB :: AB : A E,

§ 292 (if from a point without the circumference a secant and a tangent be drawn,

the tangent is a mean proportional between the whole secant and the part without the circumference).

Then AD AB : AB :: A B AE : A E.

§ 265

Since
A B= 2CB,

Cons. and

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

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

Cons. .. A D AB= A H.

Ax. 1
Also AB - A E= A B A H= H B.
Substitute these equivalents in the last proportion.
Then AH : AB :: HB : A H.
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 BA 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 : AH' : : A H' : H'B.

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

Thus AB A __B___D is 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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B 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.

From E draw the diagonals E B and E C.

From E' draw E' B', making 2 A' E' B' = 2 A E B. Also from A' draw A' B', making 2 B' A' E' = Z BAE,

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

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' Care similar, $ 280 (iwo A are similar if they have two ts of the one equal respectively to two ÉS

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

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.

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