| 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. | II.11
IV.1 | |
| Join AD and DC, and circumscribe the circle ACD about the triangle ACD. | IV.5 | |
| 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. | III.37 | |
| 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. | III.32 | |
| 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. | III.32 | |
| But the angle BDA equals the angle CBD, since the side AD also equals AB, [I. 5] so that the angle DBA also equals the angle BCD. | I.5 | |
| 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. | I.6 | |
| 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. | I.5 | |
| 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. | ||
Book IV Introduction - Proposition IV.9 - Proposition IV.11.