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PROPOSITION X. PROBLEM.
To describe an isosceles triangle, having each of the angles at the base double of the third angle.
Take any st. line AB and divide it in C,
With centre A and radius AB describe the
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,
= 4 BCD,
Hence sum of 4 s CDA, CAD ·
.. ▲ 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.
48 ADC, ACD are each=twice / DAC,
and s ADC, ACD are bisected by DB, CE,
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.
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,
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
.. 4 CKO= 2 CLO, and CK=CL,
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
.. the pentagon GHKLM is equilateral.
Thus a regular pentagon has been described about the .
To inscribe a circle in a given regular pentagon.
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,
CDE=twice / ODC,
.. 4 ABC=twice / OBC.
Hence OB bisects 4 ABC.
In the same way we can shew that OA, OE bisect
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,
So also it may be shewn that OF, OL, OK are
.. OF, OG, OH, OK, OL are all equal.
will pass through G, H, K, L,
Q. E. F.