in figure arc ab =arc ac and o is the center of the circle prove that oa is the perpendicular bisector oa bc
Answers
Given : In figure, AB ≅ AC and O is the centre of the circle.
To Prove : OA is the perpendicular bisector of BC.
Construction : Join OB and OC.
Proof :
∴ AB ≅ AC [Given]
∴ chord AB = chord AC.
[∵ If two arcs of a circle are congruent, then their corresponding chords are equal.]
∴ ∠AOB = ∠AOC ....(i) [∵Equal chords of a circle subtend equal angles at the centre]
In ∆OBC and ∆OCD,
∠DOB = ∠DOC [From (1)]
OB = OC [Radii of the same circle]
OD = OD [Common]
∴ ∆OBD ≅ ∆OCD [By SAS]
∴ ∠ODB = ∠ODC ....(ii) [By cpctc]
And BD = CD ...(ii) [By cpctc]
But ∠BDC = 1800
∴ ∠ODB + ∠ODC = 1800
⇒ ∠ODB + ∠ODB = 1800 [From equation (ii)]
⇒ 2∠ODB = 1800
⇒ ∠ODB = 900
∴ ∠ODB = ∠ODC = 900 ....(iv) [From (ii)]
So, by (iii) and (iv), OA is the perpendicular bisector of BC.
Answer:
Given : In figure, AB ≅ AC and O is the centre of the circle.
To Prove : OA is the perpendicular bisector of BC.
Construction : Join OB and OC.
Proof :
∴ AB ≅ AC [Given]
∴ chord AB = chord AC. [∵ If two arcs of a circle are congruent, then their corresponding chords are equal.]
∴ ∠AOB = ∠AOC ....(i) [∵Equal chords of a circle subtend equal angles at the centre]
In ∆OBC and ∆OCD,
∠DOB = ∠DOC [From (1)]
OB = OC [Radii of the same circle]
OD = OD [Common]
∴ ∆OBD ≅ ∆OCD [By SAS]
∴ ∠ODB = ∠ODC ....(ii) [By cpctc]
And BD = CD ...(ii) [By cpctc]
But ∠BDC = 180
∴ ∠ODB + ∠ODC = 180
⇒ ∠ODB + ∠ODB = 180 [From equation (ii)]
⇒ 2∠ODB = 180
⇒ ∠ODB = 90
∴ ∠ODB = ∠ODC = 90 ....(iv) [From (ii)]
So, by (iii) and (iv), OA is the perpendicular bisector of BC