Prove that equal chord of a circle are acute distance from centre
Answers
Answered by
2
Answer:
Step-by-step explanation:
Attachments:
Answered by
3
Given: A circle with centre O. AB and CD are two chords such that AB=CD.
To prove: The two chords are equidistant from the centre i.e. OM=ON.
Construction: Join OA, OB, OC, OD and draw OMperpendicular to AB, ON perpendicular to CD.
Proof:
Since, Perpendicular drawn from centre onto the chord bisects it.
This implies that 2AM=AB and 2CN=CD.
since,
AB=CD
AB/2=CD/2
AM=CN (1)
In triangle OAM and triangle OBN,
OA=OB(radii of the same circle)
angleOMA=angleONB (90° each)
AM=CN (from (1))
Therefore triangle OAM and triangle OBN are congruent by RHS.
This implies that OM=ON By CPCT
or distances of equal chords from the centre are equal.
Hence Proved.
To prove: The two chords are equidistant from the centre i.e. OM=ON.
Construction: Join OA, OB, OC, OD and draw OMperpendicular to AB, ON perpendicular to CD.
Proof:
Since, Perpendicular drawn from centre onto the chord bisects it.
This implies that 2AM=AB and 2CN=CD.
since,
AB=CD
AB/2=CD/2
AM=CN (1)
In triangle OAM and triangle OBN,
OA=OB(radii of the same circle)
angleOMA=angleONB (90° each)
AM=CN (from (1))
Therefore triangle OAM and triangle OBN are congruent by RHS.
This implies that OM=ON By CPCT
or distances of equal chords from the centre are equal.
Hence Proved.
Attachments:
Similar questions