Q15 of 31 Page 65

Let us write by proving the chord with the least length through any point in a circle.


To Prove CD<BG


Let, Center is O, Two chords are BG and CD, OM and OE are perpendicular bisector of BG and CD respectively.


In OME,


⇒ OE>OM


In OMB, Using Pythagoras Theorem


⇒ OB2 = OM2 + BM2 ………. (1)


In OCE


⇒ OC2 = OE2 + CE2 …………….. (2)


⇒ OC = OB


⇒ OM2 + BM2 = OE2 + CE2


⇒ OE>OM


So, BM>CE


⇒ BG>CD


As Chord goes near, its length increases.


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