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In the above figure, seg AB is a diamete...


In the above figure, seg AB is a diameter of a circle with centre P.C is any point on the circle. Seg CE `_|_` seg AB. Prove that CE is the geometric mean of AE and EB. Write the proof with the help of folloing steps:
(a) Draw ray CE. It intersects the circle at D.
(b) Show that CE = ED.
Write the result using theorem of intersection of chords insides a circle.
(d) Using CE = ED, complete the proof.

Text Solution

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In circle with centre p,AB is a diameter and seg CE `_|_` seg AB ….(Given)
`:.` seg PE `_|_` chord CD (Construction)
`:.` sec CE `~=` seg DE….(i)
(Perpendicular drawn from centre of the circle to the chord bisects the chord)
seg AB and seg CD are two chords intersecting inside the circle a point E
`:. AE xx BE = CE xx DE`
(Theorem of internal division of chords)
`AE xx BE = CE xx CE` .... (From i)
`AE xx BE = CE^(2)`
`:. CE^(2) = AE xx BE`
Hence it is proved that CE is the geometric mean of AE and EB.
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