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In the figure, two equal chords AB and CD of a circle with centre O, intersect each other at E, Prove that AD=CB |
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Answer» CF=CG `/_CFE and /_CGE` `angleCFE=angleCGE=90^o` CE=CF CF=CG `/_CFEcong/_CGE` so,FE=EG Similarly,AF=DG AF+FE=DG+GT AE=DE so,EB=EC `/_AEC and /_DEB` is similar `(AC)/(BD)=(AE)/(DE)=(CE)/(BE)=1` So, AC=BD |
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