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Hyp. The chords AB and CD meet in E.

AE X EB= CE X ED.

To prove

Hint. What is the means of proving that the product of two lines is equal to the product of two other lines ?

Ex. 632. In the diagram for Prop. XXXIII, if AE = 3, EB = 4, ED = 6, find CE.

Ex. 633. In the same diagram, if AE=a, EB=b, and ED=c, find CE.

Ex. 634. In the same diagram, if AE = 4, EB = 9, and CE = ED, find ED.

Ex. 635. If the prolongations of two chords meet without a circle, is Prop. XXXIII correct for the external segments of the chords ? Ex. 636. If two lines AB and CD intersect in E so that

AE X EB = CE X ED, then A, B, C, and D are concyclic. (See Ex. 385.)

PROPOSITION XXXIV. THEOREM 313. If, from a point without a circle, two secants are drawn, the product of one secant and its external segment is equal to the product of the other and its external segment.

Hyp. Two secants AC and AE cut the circle in B, C, and D, E, respectively.

AC X AB= AE X AD. [The proof is left to the student.]

To prove

Ex. 637. If, in the diagram for Prop. XXXIV, CB = 16, BA = 2, DA = 4 find DE.

Ex. 638. If, in the same diagram, AE should revolve about A, until AE becomes a tangent, what would AB X AC be equal to.

PROPOSITION XXXV. THEOREM

314. If, from a point without a circle, a tangent and a secant be drawn, the tangent is the mean proportional between the secant and its external segment.

D

Hyp. The tangent AB touches the O BDC in B, and the secant AD cuts the circle in C and D.

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Ex. 639. If, in the diagram for Prop. XXXV, DC = 5, AC = 4, find AB.

Ex. 640. Construct the mean proportional between two given lines by means of Prop. XXXV.

Ex. 641. Tangents to two intersecting circles, drawn from any point in the common chord produced, are equal.

Ex. 642. If the diameter of the earth is equal to 8000 mi., how far can you see from a lighthouse 100 ft. high ?

Ex. 643. If two circles are tangent externally and from any point in the common tangent secants are drawn to the two circles, the products of the secants and their external segments will be equal.

* Ex. 644. To construct a circle passing through two given points and touching a given line.

315. DEF. A straight line is said to be divided into extreme and mean ratio, when it is divided into two segments, such that the greater is the mean proportional between the smaller und the whole line. Thus, AB is divided by C in

A

с

B extreme and mean ratio, if

AB: AC = AC: CB.

H

PROPOSITION XXXVI. PROBLEM

316. To divide a line in extreme and mean ratio.

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Required. To divide AB in extreme and mean ratio.
Construction. At B, draw BC AB and I to AB.

From C as a center, with a radius CB, describe a circumference.

Draw AC meeting the circumference in D and E.
On AB, lay off AF = AD.
Then AB is divided as required.
Proof.

AE: AB= AB: AD.

AE AB: AB= AB AD: AD. (271) But

AB= DE and AD= AF.
.. AF: AB = FB: AF,

AB: AF = AF: FB.

or

Q.E.D.

317. DEF. The projection of a point upon a line is the foot of the perpendicular from the point to the line.

318. DEF. The projection of one line upon another is the length between the projections of the extremities of the first line upon

the second.

Ex. 645. If the side of an equilateral triangle equals 10 in., what is the length of the projection of one side upon another ?

Ex. 646. Find the projection of AB upon a line XY, if AB and XY include an angle of 45°, and AB = 2.

Ex. 647. Find the projection of AB upon XY, if AB = m, and the two lines include an angle of 60°.

319. Note. A abc denotes a triangle whose sides are a, b, and c. The other notations used in the following Propositions are in accordance with (230).

PROPOSITION XXXVII. THEOREM 320. In any triangle, the square of a side opposite an acute angle is equal to the sum of the squares of the other two sides diminished by twice the product of one of those sides and the projection of the other side upon it.

I

II

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To prove

Hyp. In A abc, p is the projection of b upon c, and the angle opposite a is an acute angle.

a = c

62 + c2 – 2 cp. Proof. Denote the perpendicular upon c by h. In the figure on the left

a' = l + (c-p), but

h =)? - pl, ... a? =

?

[To be completed by the student.] Similarly for Fig. II.

321. REMARK. The equation, a' = 12 + 2 – 2 cp, contains four quantities. Therefore any one of them may be found by algebraical methods if the other three are given. (Similarly in the following propositions.)

Ex. 648. The sides of a triangle are 13, 14, and 15. Find the projection of 13 upon 14.

PROPOSITION XXXVIII. THEOREM

322. In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the projection of the other side upon it.

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Hyp. In A abc, p is the projection of b upon c, and the angle opposite a is obtuse.

a = 1 + + 2cp. Proof.

a' = h' +(c +p) but

h = 52 - p2.

To prove

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