T H E E L E M E N T S |
Book III Proposition 36 | |
And the squares on FB, BD are equal to the square on FD; [I.47] therefore the rectangle AD, DC together with the square on FB is equal to the squares on FB, BD. Let the square on FB be subtracted from each; therefore the rectangle AD, DC which remains is equal to the square on the tangent DB. Being what it was required to prove. |
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