To construct an isosceles triangle having each of the angles at the base double the remaining one.

Set out any straight line *AB,* and cut it at the point *C* so that the rectangle *AB* by *BC* equals the square on *CA.* Describe the circle *BDE* with center A and radius *AB.* Fit 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.*

Join *AD* and *DC,* and circumscribe the circle *ACD* about the triangle *ACD.*

Then, since the rectangle *AB* by *BC* equals the square on *AC,* and *AC* equals *BD,* therefore the rectangle *AB* by *BC* equals the square on *BD.*

And, since a point *B* was taken outside the circle *ACD,* and from *B* the two straight lines *BA* and *BD* fall on the circle *ACD,* and one of them cuts it while the other falls on it, and the rectangle *AB* by *BC* equals the square on *BD,* therefore *BD* touches the circle *ACD.*

Since, then, *BD* touches it, and *DC* is drawn across from the point of contact at *D,* therefore the angle *BDC* equals the angle *DAC* in the alternate segment of the circle.

Since, then, the angle *BDC* equals the angle *DAC,* add the angle *CDA* to each, therefore the whole angle *BDA* equals the sum of the two angles *CDA* and *DAC.*

But the exterior angle *BCD* equals the sum of the angles *CDA* and *DAC,* therefore the angle *BDA* also equals the angle *BCD.*

But the angle *BDA* equals the angle *CBD,* since the side *AD* also equals *AB,* so that the angle *DBA* also equals the angle *BCD.*

Therefore the three angles *BDA,* *DBA,* and *BCD* equal one another.

And, since the angle *DBC* equals the angle *BCD,* the side *BD* also equals the side *DC.*

But *BD* equals *CA* by hypothesis, therefore *CA* also equals *CD,* so that the angle *CDA* also equals the angle *DAC.* Therefore the sum of the angles *CDA* and *DAC* is double the angle *DAC.*

And the angle *BCD* equals the sum of the angles *CDA* and *DAC,* therefore the angle *BCD* is also double the angle *CAD.*

But the angle *BCD* equals each of the angles *BDA* and *DBA,* therefore each of the angles *BDA* and *DBA* is also double the angle *DAB.*

Therefore the isosceles triangle *ABD* has been constructed having each of the angles at the base *DB* double the remaining one.

Q.E.F.

Euclid uses a surprising amount of the theory of circles from Book III. After drawing circle ACD, he uses III.37 to conclude from AB BC = DB^{2} that the line DB is tangent to the circle. Next, he uses III.32 to conclude that the angle BDC between the tangent DB and the chord DC equals the angle CAD which cuts off that chord.
At this point Euclid has shown that one of the two angles at |

The rest is relatively easy. First, the small triangle *BCD* is isosceles, a fact that can be seen from the following equation about angles:

There is a converse of this proposition, one the Euclid did not state. Namely, if an isosceles triangle has each base angle equal to twice the vertex angle, then the base is equal to a segment of its side so that square on the base equals the rectangle contained by the side and the remaining segment of the side. In other words, 36°-72°-72° isosceles triangles are characterized by this property.

The triangle *ABD* constructed in this proposition is one of ten sectors of a regular decagon (10-gon). Thus, it is one short step from this proposition to the construction of a regular decagon inscribed in a circle. If alternate vertices of a regular decagon are connected, then a regular pentagon is formed which is inscribed in the circle. It is unclear why Euclid did not use such a construction rather than the one he chose in the next proposition

As Euclid does, begin by cutting a straight line *AB* at the point *C* so that the rectangle *AB* by *BC* equals the square on *CA* (II.11). Otherwise said, the straight line *AB* has been cut in extreme and mean ratio at *C* so that the proportion
*AB* : *AC* = *AC* : *BC*
holds. (See VI.Def.3, VI.17, and VI.30.)
Next, construct an isosceles triangle with one side *AB,* a second side *AD* equal to side *AB,* and the base equal *BD* equal to *AC* (I.22). Then we have the proportion *AD* : *BD* = *BD* : *BC.*
Therefore, two triangles *ADB* and *DBC* have one angle equal to one angle (angle *D* of the first triangle equals angle *B* of the second) and the sides about the equal angles proportional. Therefore, by VI.6, the triangles are equiangular.
It easily follows that both triangles have their base angles each equal to twice their vertex angles.