Question

Hey Guys .... plz help me out .... If 2 equal chords intersect within a circle , prove that the segments of one chord are equal to corresponding segments of other chord .... $class 9$... ☺☺☺

Answer 1

hope it'll help you ..... :)

Answer 2

Hi there!
Here's the answer:

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Given:

Let AB and CD be the two intersecting chords intersecting at point X.

=> AB = CD

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To Prove:

Corresponding Segments are Equal
i.e., AX = DX
and BX = CX

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Proof:

We draw OM _I_ AB
& ON _I_ CD

Note that the perpendicular drawn from centre of a circle to a chord, bisects the chord.

So AM = BM = (1/2)AB
& DN = CN = (1/2)CD

As AB = CD
=> (1/2)AB = (1/2)CD

•°• AM = DN _______(1)
& MB = CN _______(2)

¶ In ∆OMX & ∆ONX,

• <OMX = <ONX (As Both are 90° )

• OX = OX (Common Angle)

• OM = ON (AB and CD are equal chords and equal chords are equidistant from the centre)

•°• ∆OMX =~ ∆ONX (As per RHS Congruence Rule)

•°• MX = NX _______(3) (CPCT)

Now,
Add (1) & (3)
=> AM + MX = DN + NX
=> AX = DX

Sub (3) from (2)
=> BM - MX = CN - MX
=> BX = CX

Therefore,
AX = BX & CX = DX

Hence Proved

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Check Attachment for the Figure

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©#£€®$

:)

Hope it helps