Book 3
15
Let EL be made equal to EH, through L let LM be drawn at right angles to EK and carried through to N, and let ME, EN, FE, EG be joined.
15
Then, since EH is equal to EL, BC is also equal to MN. [III. 14]
15
Again, since AE is equal to EM, and ED to EN, AD is equal to ME, EN.
15
But ME, EN are greater than MN, [I. 20] and MN is equal to BC; therefore AD is greater than BC.
15
And, since the two sides ME, EN are equal to the two sides FE, EG, and the angle MEN greater than the angle FEG, therefore the base MN is greater than the base FG. [I. 24]
15
But MN was proved equal to BC.
15
Therefore the diameter AD is greatest and BC greater than FG.
15
Therefore etc. Q. E. D. Of straight lines. The Greek leaves these words to be understood. Nearer to the diameter AD. As BC, FG are not in general parallel to AD, Euclid should have said nearer to the centre.
PROPOSITION 16.
16
The straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed; further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilineal angle.
16
Let ABC be a circle about D as centre and AB as diameter; I say that the straight line drawn from A at right angles to AB from its extremity will fall outside the circle.
16
For suppose it does not, but, if possible, let it fall within as CA, and let DC be joined.
16
Since DA is equal to DC, the angle DAC is also equal to the angle ACD. [I. 5]
16
But the angle DAC is right; therefore the angle ACD is also right: thus, in the triangle ACD, the two angles DAC, ACD are equal to two right angles: which is impossible. [I. 17]
16
Therefore the straight line drawn from the point A at right angles to BA will not fall within the circle.
16
Similarly we can prove that neither will it fall on the circumference; therefore it will fall outside.
