Book 4
10
But BD is by hypothesis equal to CA; therefore CA is also equal to CD, so that the angle CDA is also equal to the angle DAC; [I. 5] therefore the angles CDA, DAC are double of the angle DAC.
10
But the angle BCD is equal to the angles CDA, DAC; therefore the angle BCD is also double of the angle CAD.
10
But the angle BCD is equal to each of the angles BDA, DBA; therefore each of the angles BDA, DBA is also double of the angle DAB.
10
Therefore the isosceles triangle ABD has been constructed having each of the angles at the base DB double of the remaining one. Q. E. F.
PROPOSITION 11.
11
In a given circle to inscribe an equilateral and equiangular pentagon.
11
Let ABCDE be the given circle; thus it is required to inscribe in the circle ABCDE an equilateral and equiangular pentagon.
11
Let the isosceles triangle FGH be set out having each of the angles at G, H double of the angle at F; [IV. 10] let there be inscribed in the circle ABCDE the triangle ACD equiangular with the triangle FGH, so that the angle CAD is equal to the angle at F and the angles at G, H respectively equal to the angles ACD, CDA; [IV. 2] therefore each of the angles ACD, CDA is also double of the angle CAD.
11
Now let the angles ACD, CDA be bisected respectively by the straight lines CE, DB [I. 9], and let AB, BC, DE, EA be joined.
11
Then, since each of the angles ACD, CDA is double of the angle CAD, and they have been bisected by the straight lines CE, DB, therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another.
11
But equal angles stand on equal circumferences; [III. 26] therefore the five circumferences AB, BC, CD, DE, EA are equal to one another.
11
But equal circumferences are subtended by equal straight lines; [III. 29] therefore the five straight lines AB, BC, CD, DE, EA are equal to one another; therefore the pentagon ABCDE is equilateral.
11
I say next that it is also equiangular.
