Take any straight line AB and divide it at the point C, so that the rectangle AB, BC may be equal to the square on AC; [II. 11. from the centre A, at the distance AB describe the circle BDE, in which place the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; [IV. 1.
and join DA. The triangle ABD shall be such as is required; that is, each of the angles ABD, ADB shall be double of the third angle BAD.
Join DC; and about the triangle ACD describe the circle ACD. [IV. 5.
Then, because the rectangle AB, BC is equal to the square on AC, [Construction.
and that AC is equal to BD, [Construction.
therefore the rectangle AB, BC is equal to the square on BD.
And, because from the point B, without the circle ACD, two straight lines BCA, BD are drawn to the circumference, one of which cuts the circle, and the other meets it,
and that the rectangle AB, BC, contained by the whole of the cutting line, and the part of it without the circle, is equal to the square on BD which meets it;
therefore the straight line BD touches the circle. [I II. 37.
And, because BD touches the circle ACD, and DC is drawn from the point of contact D,
therefore the angle BDC is equal to the angle DAC in the alternate segment of the circle. [III. 32.
To each of these add the angle CDA;
therefore the whole angle BDA is equal to the two angles CDA, DAC. [Axiom 2.
But the exterior angle BCD is equal to the angles CDA DAC. [1.32.
