PROPOSITION X. PROBLEM.

To describe an isosceles triangle, having each of the angles at the base double of the third angle.

E

B

Take any st. line AB and divide it in C,
so that rect. AB, BC = sq. on AC.

With centre A and radius AB describe the

II. 11.

BDE,

IV. 1.

and in it draw the chord BD=AC; and join AD.

Then will ▲ ABD have each of the 4s at the base double

of 4 BAD.

Join CD, and about the ▲ ACD describe the ACD. IV. 5. rect. AB, BC = sq. on AC, and BD=AC,

.. rect. AB, BC = sq. on BD,

Then

= 4 BCD,

twice BAD.

Hence sum of 4 s CDA, CAD ·
.. 4 BCD =
ABD and ▲ ADB are each
L

But

L

.. ▲ ABD and ▲ ADB are each twice ▲ BAD;

and thus an isosceles ▲ ABD has nired.

been described as was

Q. E. F.

PROPOSITION XI. PROBLEM,

To inscribe a regular pentagon in a given circle.

It is required to inscribe a regular pentagon in the O. Make an isosceles ▲ FGH, having each of the ▲s at G, H double of at F.

In ABCDE inscribe a ▲ ACD equiangular to ▲ FGH, IV. 2. havings at A, C, D=the 2 s at F, G, H, respectively. Then ADC=twice DAC, and ▲ ACD=twice ▲ DAC.

Bisect the 4s ADC, ACD by the chords DB, CE.

Join AB, BC, DE, EA.

Then will ABCDE be a regular pentagon.

For

48 ADC, ACD are each=twice / DAC,

and s ADC, ACD are bisected by DB, CE,
.. ¿s ADB, BDC, DAC, ECD, ACE, are all equal;

and.. arcs AB, BC, CD, DE, EA are all equal; and.. chords AB, BC, CD, DE, EA are all equal. Hence, the pentagon ABCDE is equilateral. Again, ·.· arc CD=arc AB,

adding to each arc AED, we have

arc AEDC=arc BAED,

Similarly, s CDE, DEA, EAB each= ▲ ABC.

Hence, the pentagon ABCDE is equiangular.

III. 26.

III. 29.

III. 27.

Thus a regular pentagon has been inscribed in the .

Ex. Shew that CE is parallel to BA.

Q. E. F.

PROPOSITION XII. PROBLEM.

To describe a regular pentagon about a given circle.

It is required to describe a regular pentagon about the ○. Let the angular pts. of a regular pentagon inscribed in the be at A, B, C, D, E,

so that the arcs AB, BC, CD, DE, EA are all equal.

Through A, B, C, D, E draw GH, HK, KL, LM, MG tangents to the ;

take the centre O, and join OB, OK, OC, OL, OD.

Then in AS OBK, OCK,

** OB=OC, and OK is common, and KB=KC,

:. 2 BK0= 2 CKO, and ▲ BOK= 4 COK,

L

I. E. Cor.

that is, ▲ BKC=twice 4 CKO, and ▲ BOC=twice 4 COK.

So also, ▲ DLC=twice ▲ CLO, and ▲ DOC=twice ▲ COL.

Now arc BC-arc CD,

..L BOC= L DOC,

and .. 4 COK= ▲ COL.

Hence in As OCK, OCL,

:: 2 COK= 2 COL, and rt. 4 OCK=rt. 2 OCL, and OC is

common,

.. 4 CKO= 2 CLO, and CK=CL,

I. B.

and .. 2 HKL= 2 MLK, and KL=twice KC. Similarly it may be shewn that 2 8 KHG, HGM, GML each = 1 HKL,

.. the pentagon GHKLM is equiangular.

And since it has been shewn that KL=twice KC,

and it can be shewn that HK=twice KB,

In like manner it may be shewn that HG, GM, ML, each

= KL,

.. the pentagon GHKLM is equilateral.

Thus a regular pentagon has been described about the .

QE. F.

To inscribe a circle in a given regular pentagon.

A

K

H

Let ABCDE be the given regular pentagon.

It is required to inscribe a ○ in the pentagon. Bisect 48 BCD, CDE by the st. lines CO, DO, meeting in O. Join OB, OA, OE.

Then, in as BCO, DCO,

• BC=DC, and CO is common, and ▲ BCO= ▲ DCO,

..L OBC= 4 ODC.

Then, ABC= ▲ CDE,

and

CDE=twice / ODC,

.. 4 ABC=twice / OBC.

Hence OB bisects 4 ABC.

In the same way we can shew that OA, OE bisect
the S BAE, AED.

I. 4. Hyp.

Draw OF, OG, OH, OK, OL1 to AB, BC, CD, DE, EA. Then, in AS GOC, HOC,

::: 4 GCO= 4 HCO, and ▲ OGC= ▲ OHC,

and OC is common,

.. OG=OH.

So also it may be shewn that OF, OL, OK are

.. OF, OG, OH, OK, OL are all equal.
described with centre O and radius OF

Hence a

will pass through G, H, K, L,
and will touch the sides of the pentagon,
...the 4s at F, G, H, K, L are rt. 4 s.
Thus a will be inscribed in the pentagon.

I. 26.

III. 16.

Q. E. F.