AC and BD are chords of a circle which bisect each other. Prove that:(i) AC and BD are diameters.
(ii) ABCD is a rectangle.
Given: AC and BD are chords of a circle which bisect each other at O. (say).To Prove: (i) AC and BD are diameters(ii) ABCD is a rectangle
Construction: Join AB, BC, CD, and DA.Proof: (i) In ∆OAB and ∆OCD,OA = OC| ∵ O is the mid-point of AC∠AOB = ∠COD| Vertically opposite anglesOB = OD| ∵ O is the mid-point of BD∴ ∆OAB ≅ ∆OCD| SAS congruence rule∴ AB = CD | C.P.C.T
| If two chords of a circle are equal, then their corresponding arcs are congruentIn ∆OAD and ∆OCB,OA = OC| ∵ O is the mid-point of AC∠AOD = ∠COB| Vertically opposite anglesOD = OB| ∵ O is the mid-point of BD∴ ∆OAD = ∆OCB| SAS congruence rule∴ AD = CB | C.P.C.T.
| If two chords of a circle are equal, then their corresponding arcs are congruentFrom (1) and (2),
| A parallelogram with one of its angles 90° is a rectangle
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If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel.
Solution
Given: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at points L and M. To prove: AB || CD Proof: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at L and M. Then OL ⊥ AB Also, OM ⊥ CD ∴ ∠ ALM = ∠ LMD = 90 o Since alternate angles are equal, we have:AB|| CD
Mathematics
Secondary School Mathematics IX
Standard IX
6