In fig 4.43 two equal chords AB and CD of a circle with centre O. When produced meet at point E prove that BE = DE and AE = EC.
(Hint: Draw OL $⊥$ AB and ON $⊥$ CD, join OE)

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Solution

$Draw OM ⊥ AB and ON ⊥ CD . We know that perpendicular drawn from the centre of the circle bisects the chord . So, AM = BM and CN = DN . AB = CD (given) So, BM = ND . . . (1) Now, in △ OME and △ ONE, OE = OE (common) ON = OM (e qual chords are equidistan t from the circle) ∠ ONE = ∠ OME = 90 ° △ OME ≅ △ ONE (by RHS congruency) ME = NE (c . p . c . t) Now, subtracting BM from both sides, we get : ME - BM = NE - BM \Rightarrow ME - BM = NE - ND (u sing eq . 1) \Rightarrow BE = DE . . . (2) AB = CD Adding BE on both sides, we get : AB + BE = CD + BE \Rightarrow AB + BE = CD + DE (from eq . 2) \Rightarrow AE = EC$

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In fig 4.43 two equal chords AB and CD of a circle with centre O. When produced meet at point E prove that BE = DE and AE = EC. (Hint: Draw OL ⊥ AB and ON ⊥CD, join OE)

In fig 4.43 two equal chords AB and CD of a circle with centre O. When produced meet at point E prove that BE = DE and AE = EC.
(Hint: Draw OL $⊥$ AB and ON $⊥$ CD, join OE)