PROPOSITION XXIII. PROBLEM. 304. To find a fourth proportional to three given straight lines.

Let the three given lines be m, n, and o.

It is required to find a fourth proportional to m, n, and o.

Take A B equal to n. Draw the indefinite line A R, making any convenient z with A B.

On A R take A C=m, and CS=0.

Draw C B.

From S draw SF || to C B, to meet A B produced at F.

BF is the fourth proportional required.

For, AC : AB ::CS : BF,

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

proportionally).

Substitute m, n, and o for their equals A C, A B, and C S.

Then

m :n ::0 : BF.

Q. E. F.

PROPOSITION XXIV. PROBLEM. 305. To find a third proportional to two given straight lines.

Let A B and AC be the two given straight lines.

It is required to find a third proportional to A B and A C.
Place A B and A C so as to contain any convenient 2.
Produce A B to D, making B D= AC.

Join BC.
Through D draw D E Il to B C to meet A C produced at E.

C E is a third proportional to A B and AC.

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For,

AB AC

For,
BDCE'

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

proportionally).

Substitute, in the above equality, A C for its equal B D;

Then

A B AC
A C = CE'

AB : AC :: AC : CE.

Q. E. F.

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

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 I CH.
CH is a mean proportional between m and n.

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

§ 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 : CB,

$ 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 A C and C B their equals m and n.
Then
m : C H :: C H : 0.

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.

C

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
CA CE

§ 275 AF – EB (a line drawn through two sides of a A ll to the third side divides those sides

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

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=LCAE,

§ 70 (being ext.-int. 6). ZABF= L B A E,

§ 68 (being alt.-int. 6). ...ZCA E= Z B A E.

Ax. 1

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.

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 : AE,

§ 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 : A B :: A B – AE : A E.

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