270. THEOREM. If the diagonals drawn from one vertex in each of two polygons divide them into the same number of triangles, similar each to each and similarly placed, then the two polygons are similar.

Given the diagonals drawn from the vertices A and A' in the polygons P and P', forming the same number of triangles in each, such that I~ ^I', ▲ II ~ ^ II', ▲ III ~ ▲ III'.

271. THEOREM. State the converse of the preceding theorem, and give the proof in full detail.

Make an outline of all the steps in the proofs of these two theorems.

272. THEOREM. If through a fixed point within a circle any number of chords are drawn, the product of the segments of one chord is equal to the product of the segments of any other.

B

E

Given C with any two chords AB and DE intersecting in P.

To prove that AP PB EP. PD.

Proof : Draw EB and DA.

=

Which are corresponding angles and which are corresponding sides?

It follows from this theorem that if a chord AB is made to swing around the fixed point P, the product AP·PB does not change, that is, it is constant.

1. Which chord through a point is bisected by the diameter through that point? Why?

2. Through a given point within a circle which chord is the shortest? Why?

3. The product AP. PB is the area of the rectangle whose base and altitude are the segments of AB. (See § 307.)

Note that this area is constant as the chord swings about the point P as a pivot.

274. Definition.

If a secant of a circle is drawn from a

point P without it, meeting the circle in the points A and B, then PB is called the whole secant and PA the external segment, provided A lies between B and P.

275. THEOREM. If from a fixed point outside a circle any number of secants are drawn, the product of one whole secant and its external segment is the same as that of any whole secant and its external segment.

Given secants PB and PE drawn from a point P.

To prove that PA · PB = PD · PE.

Proof : In the figure show that ▲ PDB ~ \ PAE.
Complete the proof.

276.

EXERCISES.

1. A point P is 8 inches from the center of a circle whose radius is 4. Any secant is drawn from P, cutting the circle. Find the product of the whole secant and its external segment.

2. From the same point without a circle two secants are drawn. If one whole secant and its external segment are 14 and 5 respectively and the other external segment is 7, find the other whole

secant.

3. Two chords intersect within a circle. The segments of one are m and n and one segment of the other is p. Find the remaining segment.

277. THEOREM. If a tangent and a secant meet outside a circle, the square on the tangent is equal to the product of the whole secant and

D

A

1. If a square is constructed on PD as a side, and a rectangle with PB as base and PA as altitude, compare their areas as the secant revolves about P as a pivot.

2. Show that the theorem in § 277 may be obtained as a direct consequence of that in § 275 by supposing one secant to swing about P as a pivot till it becomes a tangent.

3. A point P is 10 inches from the center of a circle whose radius is 6 inches. Find the length of the tangent from P to the circle.

4. The length of a tangent from P to a circle is 7 inches, and the external segment of a secant is 4 inches. Find the length of the whole secant.

5. What theorems are included in the following statement : “From a point P in a plane a line is drawn cutting a circle in A and B. Then the product PA PB is the same for all such lines" "?

6. In a circle of radius 10 a point P divides a chord into two segments 4 and 6. How far from the center is P? Use Ex. 5.

SUGGESTION.

7. In two similar polygons two corresponding sides are 3 and 7. If the perimeter of the first polygon is 45, what is the perimeter of the second?

8. The perimeters of two similar polygons are 32 and 84. A side of the first is 11. What is the corresponding side of the second polygon?

In a right triangle ABC, right-angled

is called the sine of A

and is written sin A.

If any other point B' be taken on the hypotenuse or the hypotenuse extended, and a perpendicular B'C' be let fall to AC, A C'B' CB

then

AB' AB

(Why?)

B

B

B

Sin A

Likewise in ▲ AB"C", in which AB" = 1 unit, we have

Hence, in a right triangle whose hypotenuse is unity, the length of the side opposite an acute angle is the sine of that angle.

280. THEOREM.

The ratio of the sides opposite two acute angles of a triangle is equal to the ratio of the

sines of these angles.

Given ▲ ABC with A and B both acute angles.

NOTE. The definitions of § 279 and the theorem of § 280 are given here for acute angles only. In trigonometry, where the subject is studied in full detail, they are extended to apply to any angles whatever. Other ratios called cosines, tangents, etc., are also introduced.