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.

:P1E43AB7C 1.png



Note that OFD and OKD are right angles.


Construct FK and label the four angles.

:P1E43AB7C 2.png



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

:P1E43AB7C 3.png


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.


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