Q14 of 31 Page 65

I prove that, among two chords of a circle the length of the nearer to centre is greater than the length of the other.


To Prove: DH>EG


Construction: B is the center of the circle, CD and EF are the chords, BH and BG are the perpendicular bisector of CD and EF respectively.


Given BH<BG As it given, CD is nearer to center than EF.


In BDH


⇒ BD2 = BH2 + DH2 …………… (1)


In BEG


⇒ BE2 = BG2 + EG2 ………… (2)


⇒ BD = BE


From Eq1 and Eq2


⇒ BH2 + DH2 = BG2 + EG2


⇒ BH<BG


⇒ BH2<BG2


⇒ BH2-BG2<0


From Eq.3


⇒ BH2 + DH2 = BG2 + EG2


⇒ BH2-BG2 = EG2-DH2


⇒ EG2-DH2<0


⇒ EG2<DH2


⇒ EG<DH


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