PROPOSITION 10.
To construct an isosceles triangle having each of the angles at the base double of the remaining one.
Let any straight line
AB be set out, and let it be cut at the point
C so that the rectangle contained by
AB,
BC is equal to the square on
CA; [
II. 11] with centre
A and distance
AB let the circle
BDE be described, and let there be fitted in the circle
BDE the straight line
BD equal to the straight line
AC which is not greater than the diameter of the circle
BDE. [
IV. 1]
Let
AD,
DC be joined, and let the circle
ACD be circumscribed about the triangle
ACD. [
IV. 5]
Then, since the rectangle
AB,
BC is equal to the square on
AC, and
AC is equal to
BD, therefore the rectangle
AB,
BC is equal to the square on
BD.
And, since a point
B has been taken outside the circle
ACD, and from
B the two straight lines
BA,
BD have fallen on the circle
ACD, and one of them cuts it, while the other falls on it, and the rectangle
AB,
BC is equal to the square on
BD,
therefore BD touches the circle ACD. [III. 37]
Since, then,
BD touches it, and
DC is drawn across from the point of contact at
D, therefore the angle
BDC is equal to the angle
DAC in the alternate segment of the circle. [
III. 32]
Since, then, the angle
BDC is equal to the angle
DAC, let the angle
CDA be added to each; therefore the whole angle
BDA is equal to the two angles
CDA,
DAC.
But the exterior angle
BCD is equal to the angles
CDA,
DAC; [
I. 32] therefore the angle
BDA is also equal to the angle
BCD.
But the angle
BDA is equal to the angle
CBD, since the side
AD is also equal to
AB; [
I. 5] so that the angle
DBA is also equal to the angle
BCD.
Therefore the three angles
BDA,
DBA,
BCD are equal to one another.
And, since the angle
DBC is equal to the angle
BCD,
the side BD is also equal to the side DC. [I. 6]
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.
But the angle
BCD is equal to the angles
CDA,
DAC; therefore the angle
BCD is also double of the angle
CAD.
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.
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.