2. Let EO be equal to FO, it is required to prove AB equal to CD. The same construction being made, we have, as before, AE2 + EO2 equal to CF2 + FO2; but EO2 is equal to FO2 (hyp.). Hence AE2 is equal to CF2, and AE is equal to CF; but AB is double of AE, and CD double of CF. Therefore AB is equal to CD.
Exercise.
If a chord of given length slide round a fixed circle—1. the locus of its middle point is a circle; 2. the locus of any point fixed in the chord is a circle.
PROP. XV.—Theorem.
The diameter (AB) is the greatest chord in a circle; and of the others, the chord (CD) which is nearer to the centre is greater than (EF) one more remote, and the greater is nearer to the centre than the less.
Dem.—1. Join OC, OD, OE, and draw the perpendiculars OG, OH; then because O is the centre, OA is equal to OC [I., Def. xxxii.], and OB is equal to OD. Hence AB is equal to the sum of OC and OD; but the sum of OC, OD is greater than CD [I. xx.]. Therefore AB is greater than CD.
2. Because the chord CD is nearer to the centre than EF, OG is less than OH; and since the triangles OGC, OHE are right-angled, we have OC2 = OG2 + GC2, and OE2 = OH2 + HE2; therefore OG2 + GC2 = OH2 + HE2; but OG2 is less than OH2; therefore GC2 is greater than HE2, and GC is greater than HE, but CD and EF are the doubles of GC and HE. Hence CD is greater than EF.
3. Let CD be greater than EF, it is required to prove that OG is less than OH.
As before, we have OG2 + GC2 equal to OH2 + HE2; but CG2 is greater than EH2; therefore OG2 is less than OH2. Hence OG is less than OH.