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let F be the center of the circle ABC;
let FB be joined;
therefore the angle FBD is right. [III.18]
And, since AC has been bisected at F, and CD is added to it,
the rectangle AD, DC together with the square on FC is equal to the square on FD. [II.6]
But FC is equal to FB; therefore
the rectangle AD, DC together with the square on FB is equal to the square on FD.
