Theorem: The chords of a circle equidistant from the centre of a circle are congruent. Given In a circle with centre O seg OP I chord AB seg OQ I chord CD and OP=OQ : To prove: chord AB chord CD Construction: Draw seg OA and seg OD. Proof : (Complete the proof by filling in the gaps.)
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Verified answer
Answer:
Given : A circle have two equal chords AB and CD.
AB=CD and OM perpendicular to AB, ON perpendicular to CD
To Prove : OM=ON
Proof : AB=CD (Given)
∵ the perpendicular drawn from the centre of a circle to bisect the chord
∴
2
1
AB=
2
1
CD
⇒BM=DN
In △OMB and △OND
∠OMB=∠OND=90
0
[Given]
OB=OD [Radii of same circle]
Side BM= Side DN [Proved above]
∴△OMB≅△OND [By R.H.S.]
∴OM=ON [By C.P.C.T]
Proof :- (Complete the proof by filling in the gaps.)
In right angled ∆OPA and right ∆OQD.
hypotenuse OA = hypotenuse OD ------ ( Radii of the sam circle )
seg OP = seg OQ -----( given )
∴ ∆OPA = ∆OQD ------- ( Hypotenuse side test of congruence)
∴ seg AP = seg QD ---- ( c.s.c.t. )
∴ AP = QD
But AP = ½ AB and DQ = ½ CD ------(The perpendicular drawn the centre of a circle to its chord bisects the)
and AP = QD
∴ AB = CD
∴ seg AB = seg CD.
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