Again, let DCA not be through the centre of the circle ABC;
let the centre E be taken, and from E let EF be drawn perpendicular to AC;
let EB, EC, ED be joined.
Then the angle EBD is right. [III. 18]
And, since a straight line EF through the centre cuts a straight line AC not through the centre at right angles,
it also bisects it; [III. 3]
therefore AF is equal to FC.
Now, since the straight line AC has been bisected at the point F, and CD is added to it, the rectangle contained by AD, DC together with the square on FC is equal to the square on FD. [II. 6]
Let the square on FE be added to each;
therefore the rectangle AD, DC together with the squares on CF, FE is equal to the squares on FD, FE.
But the square on EC is equal to the squares on CF, FE, for the angle EFC is right; [I. 47]
and the square on ED is equal to the squares on DF, FE;
therefore the rectangle AD, DC together with the square on EC is equal to the square on ED.
And EC is equal to EB;
therefore the rectangle AD, DC together with the square on EB is equal to the square on ED.
But the squares on EB, BD are equal to the square on ED,
for the angle EBD is right; [I. 47]
therefore the rectangle AD, DC together with the square on EB is equal to the squares on EB, BD.
Let the square on EB be subtracted from each;
therefore the rectangle AD, DC which remains is equal to the square on DB.