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with the same construction,
we can prove similarly that
BC is equal to BG;
so that
the angle at C is also equal to the angle BGC. [I.5]
But
the angle at C is not less than a right angle;
therefore
neither is the angle BCG less than a right angle.
Thus
in the triangle BGC the two angles are not less than two right angles: which is impossible. [I.17]
