Solution:As, AB=CDSo, OL=OM (Chords are equidistant from the center if and only if their lengths are equal )Also, OL and OM bisects AB and CD respectively Theorem:The perpendicular from the center of a circle to a chord bisects the chord12AB=12CD⇒AL=CM (1)Join OE So, now in tri OLE and tri OME:OL=OM(from above)OE (Common)Angle OLE=Angle OME=90° So, RHS criterion both are triangles are congruent Therefore, LE=ME (CPCT)⇒LE+AL=ME+AL⇒LE+AL=ME+CM (From (i))⇒AE=CE

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Solution : As, AB = CD So, OL = OM (Chords are equidistant from the center if and only if their lengths are equal .) Also, OL and OM bisects AB and CD respectively . Theorem : The perpendicular from the center of a circle to a chord bisects the chord \frac{1}{2} AB = \frac{1}{2} CD \Rightarrow AL = CM . . . . (1) Join OE . So, now in tri . OLE and tri . OME : OL = OM (from above) OE (Common) Angle OLE = Angle OME = 90 ° So, RHS criterion both are triangles are congruent . Therefore, LE = ME (CPCT) \Rightarrow LE + AL = ME + AL \Rightarrow LE + AL = ME + CM (From (i)) \Rightarrow AE = CE

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