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If a line intersects two concentric circles (circles with the same
centre) with centre O at A, B, C and D, prove that AB = CD.
Draw a perpendicular OM on line AD.
BC is the chord of the smaller circle and AD is the chord of the bigger circle.
We know that perpendicular drawn from the centre of the circle bisects the chord.
BM = MC ... (1)
And, AM = MD ... (2)
On subtracting equation (1) from (12, we get
AM − BM = MD − MC
AB = CD
Two equal chords AB and CD of circle with center O, when produced
meet at a point E. Prove that BE = DE and AE = CE.Marks:5
Let OL and OM be two perpendiculars from centre O to chords AB and CD
respectively. So L and M are the mid points of chords AB and CD.
Now AB = CD (Given) ...(1)
AB/2 = CD/2 or LB = MD
In OLE and OME
OL = OM (Equal Chords are equidistant from the centre)
OLE = OME (90° each)
OE = OE (common side)
So by RHS
This gives, LE = ME ( by CPCT)
Here LE = LB + BE and ME = MD + DE which gives,
LB + BE = MD + DE
But LB = MD (proved above)
So, BE = DE ... (2)
Adding (1) and (2), we get
AB + BE = CD + DE
Therefore, AE = CE
If two equal chords of a circle intersect within the circle at a point, then prove that the segments of one chord are equal to the corresponding segments of the other chord.Marks:5
Given: A circle with centre O and chord AB = chord CD.
Chords AB and CD intersect each other at E.
To Prove: AE = CE and EB = ED
In the given figure, O is the centre of a circle; OA=10 cm, OC=6 cm. Find the length of AB.