**Problem 2.1.1b**

Given a circle O with chords AB and CD such that AB < CD. Prove OE > OF where OE is the distance along a perpendicular from the center to chord AB and OF is the distance to CD.

We know that E is the midpoint of AB and F is the midpoint of CD.

Construct chord DH such that AB = DH. Let K be the midpoint of DH and OK the perpendicular segment to DH. By the construction, we have AB congruent to DH and OE congruent to OK.

Note that OFD and OKD are right angles.

Construct FK and label the four angles.

Now FD = 1/2 CD and KD = 1/2 HD = 1/2 AB. So FD > KD. Therefore by Theorem 1.13 or 1.14, we have

From this OF is opposite the smaller angle and OK is opposite the larger angle in triangle FOK.

Therefore OF < OK, or to be proved, OF < OE.