Hence KL is double of KC; similarly HK is double of KB.

And since KC = KB,

.. KL HK.

=

III. 17, Cor.

In the same way it may be shewn that every two consecutive sides are equal;

..the pentagon GHKLM is equilateral.

Again, it has been proved that the FKC = the FLC, and that the HKL, KLM are respectively double of these angles :

In the same way it may be shewn that every two consecutive angles of the figure are equal;

..the pentagon GHKLM is equiangular. .. the pentagon is regular, and it is circumscribed about

the ABCD.

Q.E.F.

COROLLARY. Similarly it may be proved that if tangents are drawn at the vertices of any regular polygon inscribed in a circle, they will form another regular polygon of the same species circumscribed about the circle.

[For Exercises see p. 293.]

PROPOSITION 13. PROBLEM.

To inscribe a circle in a given regular pentagon.

M

E

H

K

Let ABCDE be the given regular pentagon.

It is required to inscribe a circle within the figure ABCDE.

Construction.

Bisect two consecutive

CF and DF which intersect at F.

Join FB;

and draw FH, FK perp. to BC, CD.

BCD, CDE by

I. 9.

I. 12.

But the

... also the

CDF is half an angle of the regular pentagon : CBF is half an angle of the regular pentagon : that is, FB bisects the

ABC.

So it may be shewn that if FA, FE were joined, these lines would bisect the

at A and E.

Similarly if FG, FM, FL be drawn perp. to BA, AE, ED, it may be shewn that the five perpendiculars drawn from F to the sides of the pentagon are all equal,

M

E

H

K

With centre F, and radius FH, describe a circle; this circle must pass through the points H, K, L, M, G; and it will be touched at these points by the sides of the pentagon, for the at H, K, L, M, G are rt. 2o. Constr. ..the HKLMG is inscribed in the given pentagon. Q.E.F.

COROLLARY. The bisectors of the angles of a regular pentagon meet at a point.

NOTE. In the same way it may be shewn that the bisectors of the angles of any regular polygon meet at a point. [See Ex. 1, p. 294.]

[For Exercises on Regular Polygons see p. 293.]

MISCELLANEOUS EXERCISES.

1. Two tangents AB, AC are drawn from an external point A to a given circle: describe a circle to touch AB, AC and the convex arc intercepted by them on the given circle.

2. ABC is an isosceles triangle, and from the vertex A a straight line is drawn to meet the base at D and the circumference of the circumscribed circle at E: shew that AB is a tangent to the circle circumscribed about the triangle BDE.

3. An equilateral triangle is inscribed in a given circle: shew that twice the square on one of its sides is equal to three times the area of the square inscribed in the same circle.

4. ABC is an isosceles triangle in which each of the angles at B and C is double of the angle at A; shew that the square on AB is equal to the rectangle AB, BC with the square on BC

To circumscribe a circle about a given regular pentagon.

Let ABCDE be the given regular pentagon.

It is required to circumscribe a circle about the figure ABCDE.

Bisect the BCD, CDE by CF, DF, inter

Construction.

secting at F.

But the

... also the

=

I. 9.

Hyp.

DCF;

Constr.

CDF.

I. 4.

CDF is half an angle of the regular pentagon : CBF is half an angle of the regular pentagon : that is, FB bisects the

ABC.

So it may be shewn that FA, FE bisect the 4 at A and E.

Now the FCD, FDC are each half an angle of the given regular pentagon;

... the FCD = the FDC,

... FC=FD.

IV. Def. 2.
I. 6.

Similarly it may be shewn that FA, FB, FC, FD, FE are all equal.

With centre F, and radius FA, describe a circle : this circle must pass through the points A, B, C, D, E, and therefore is circumscribed about the pentagon. Q.E.F.

NOTE. In the same way a circle may be circumscribed about any regular polygon.

To inscribe a regular hexagon in a given circle.

B

H

Let ABDF be the given circle.

ABDF.

ABDF; III. 1.

It is required to inscribe a regular hexagon in the
Construction. Find G the centre of the

and draw a diameter AGD.

With centre D, and radius DG, describe the O EGCH. Join CG, EG, and produce them to cut the Oce of the given circle at F and B.

Join AB, BC, CD, DE, EF, FA.

Then ABCDEF shall be the required regular hexagon.

Proof. Now GE=GD, being radii of the O ACE;
and DG = DE, being radii of the

=

EHC:

... GE, ED, DG are all equal, and the ▲ EGD is equilateral. Hence the EGD = one-third of two rt. angles. I. 32. Similarly the DGC = one-third of two rt. angles. But the EGD, DGC, CGB together two rt. angles; I. 13. .. the remaining 4 CGB = one-third of two rt. angles. .. the three EGD, DGC, CGB are equal to one another. And to these angles the vert. opp. 4 BGA, AGF, FGE are respectively equal:

.. the EGD, DGC, CGB, BGA, AGF, FGE are all equal; .. the arcs ED, DC, CB, BA, AF, FE are all equal: III. 26. .. the chords ED, DC, CB, BA, AF, FE are all equal: III. 29. .. the hexagon is equilateral. Again the arc FA = the arc DE:

to each of these equals add the arc ABCD;

then the arc FABCD the arc ABCDE:

=

Proved.

hence the angles at the Oce which stand on these equal arcs are equal.