26. Trisect a given triangle by drawing straight lines through any given point on one of the sides.

C

27. Bisect a given triangle by a straight

line drawn parallel to one of the sides.

DB a median, ADCB must equal

ANALYSIS. - If EF is the required line, and

Why? Hence CE × CF = CD × CB. Why? A

ECF.

CD CE

=

,

that is, CE is the mean proportional between CD and CA.

CE CA

28. Bisect any quadrilateral by a straight line drawn through one of the vertices.

A

ANALYSIS. - Let A be the vertex. If E is the middle point of BD, then AE and CE divide the quadrilateral into two equal parts.

D

Hence it remains to construct a triangle equal to AAEC, having for base AC and a vertex in one of the sides of the quadrilateral.

29. Trisect any given quadrilateral by straight lines drawn through any vertex.

30. Divide a given parallelogram into any required number of equal parts by straight lines drawn from one of its vertices.

31. Divide a given parallelogram into four equal parts by drawing straight lines through any given point on one of its sides.

SUGGESTION. - Let P be the given point. Divide AB into four equal parts. Draw EF and GH parallel to AD. Bisect EF at R and GH at S. Having drawn PM and PN, construct PO by Ex. 28.

DMF ΗΝ

R

C

0

S

CHAPTER XI

REGULAR POLYGONS AND CIRCLES

239. Regular polygons. A regular polyyon (See § 79) is equiangular and equilateral.

Regular polygons occur in nature, as in honeycomb. They are used in designs, as in tile floors, etc. The figure shows regular hexagons in a lace pattern.

240. Theorem. A circle may be circumscribed about any regular polygon.

E

Conclusion. A circle may be circumscribed about ABCD ....

2. A circle may be drawn through A, B, and C. 3. Let its center be 0. Draw OA, OB, OC, OD.

§ 152

4. OB = 0С.

§ 151

5.... 1 = ∠ 2.

§ 70

6. ∠ABC = ∠ BCD and AB = CD.

7. ... 3 = 24.

8. ... ΔΑOB ≃ △ COD.

9. ... OA = OD, and OD is a radius.

10. ... the circle passes through D.

11. Similarly, it goes through E, etc., and is circumscribed.

Def. reg. poly.

Ax. III

§ 63

Def. congruence

Def. radius

241. Theorem. A circle may be inscribed in any regular polygon.

F

E

D

P

C

A--M---B

Hypothesis. ABCD ... is any regular polygon. Conclusion. A circle may be inscribed in ABCD .... Suggestions. A circle with center O may be circumscribed about ABCD.... Why? Draw OMLAB, ON⊥ BC, OPᅩ CD, etc. Prove OM = ON = OP = etc. Now draw a circle with center O and radius OM. Prove AB, BC, CD, etc., tangents to it.

242. Corollary 1. — The inscribed and circumscribed circles of a regular polygon are concentric.

This was shown in the proof of § 241.

243. Parts of a regular polygon. - The center of a regular polygon is the common center of the inscribed and circumscribed circles.

The radius of a regular polygon is the radius of the circumscribed circle.

The apothem of the polygon is the radius of the inscribed circle.

The angle at the center of a regular polygon is the angle between the radii drawn to the extremities of any side.

244. Corollary 2. — The angle at the center of any regular polygon is equal to 360° divided by the number of sides. Why?

EXERCISES

1. How many degrees are there in each angle of a regular polygon of 10 sides? Of 16 sides.? Of 20 sides?

2. How many degrees are there in the angle at the center of a regular pentagon? Of a regular hexagon? Of a regular octagon? Of a regular decagon?

3. What is the relation between the angle at the center of a regular polygon and the angle at a vertex? Give proof.

4. Prove that if all diagonals are drawn from any vertex of a regular polygon, they divide the angle at that vertex into equal parts.

5. Prove that the radius drawn to any vertex of a regular polygon bisects the angle at that vertex.

6. Prove that the apothem of a regular polygon bisects any side to which it is drawn.

7. Prove that the apothern of an equilateral triangle is equal to one half of the radius.

8. Prove that the apothem of an equilateral triangle is equal to one third of the altitude of the triangle.

245. Theorem. - If a circle is divided into any number of equal arcs, the chords of these arcs form a regular inscribed polygon of that number of sides.

many sides as there are arcs AB, etc.

Suggestions. Why are the sides equal? = arc CEA etc. Then ∠A = ∠B, etc. the proof.

Prove arc BDF
Why? Write

246. Corollary 1. - An equilateral polygon inscribed in a circle is regular.

The proof is left to the student.

247. Corollary 2. - If from the vertices of a regular inscribed polygon line-segments are drawn to the middle points of the arcs subtended by the sides, they form another regular inscribed polygon of twice as many sides.

The proof is left to the student.

248. Theorem. - If a circle is divided into any number of equal arcs, the tangents at the points of division form a regular circumscribed polygon of that number of sides.

Hypothesis. MN = NP = PQ = etc., and AB, BC, CD, etc., are tangents at M, N, P, etc., respectively.

Conclusion. ABCD

...

is a regular polygon, having as

many sides as there are arcs MN, etc.

Now ∠BMN = ∠ MNB prove that MBN,

Suggestions. Draw MN, NP, etc. = ∠ CNP = etc. Why? Then ANCP, etc., are congruent isosceles triangles, and hence that MB = BN = NC = CP = etc. Then prove AB = BC

=

CD = etc. Why does ∠A = ∠ B = ∠ C = etc. ?

Write the proof in full.