Kitap 12
17
Then, since AX is at right angles to the plane of the quadrilateral KBPS, therefore it is also at right angles to all the straight lines which meet it and are in the plane of the quadrilateral. [XI. Def. 3]
17
Therefore AX is at right angles to each of the straight lines BX, XK.
17
And, since AB is equal to AK, the square on AB is also equal to the square on AK.
17
And the squares on AX, XB are equal to the square on AB, for the angle at X is right; [I. 47] and the squares on AX, XK are equal to the square on AK. [id.]
17
Therefore the squares on AX, XB are equal to the squares on AX, XK.
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Let the square on AX be subtracted from each; therefore the remainder, the square on BX, is equal to the remainder, the square on XK; therefore BX is equal to XK.
17
Similarly we can prove that the straight lines joined from X to P, S are equal to each of the straight lines BX, XK.
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Therefore the circle described with centre X and distance one of the straight lines XB, XK will pass through P, S also, and KBPS will be a quadrilateral in a circle.
17
Now, since KB is greater than WV, while WV is equal to SP, therefore KB is greater than SP.
17
But KB is equal to each of the straight lines KS, BP; therefore each of the straight lines KS, BP is greater than SP.
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And, since KBPS is a quadrilateral in a circle, and KB, BP, KS are equal, and PS less, and BX is the radius of the circle, therefore the square on KB is greater than double of the square on BX.
17
Let KZ be drawn from K perpendicular to BV.
