Prove that the diameter of a circle that bisects a chord also bisects the angle subtended by the chord at the centre of the circle.
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Given : A circle with centre O of diameter AB bisects a chord CD at e (i.e, CE = ED)
To prove: diameter AB also bisects angle (i.e, angle COE = angle DOE)
Proof:
In triangles COE and DOE,
OE is common to both,
CE = ED [given]
OC = OD [radii of circle]
now,
triangle COE is congruent to triangle DOE
therefore,
angle COE = angle DOE [corresponding angles]
Hence proved.
To prove: diameter AB also bisects angle (i.e, angle COE = angle DOE)
Proof:
In triangles COE and DOE,
OE is common to both,
CE = ED [given]
OC = OD [radii of circle]
now,
triangle COE is congruent to triangle DOE
therefore,
angle COE = angle DOE [corresponding angles]
Hence proved.
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Answer:
Form figure: PQ is a diameter of circle which bisects the chord AB at C. (Given) To Prove: PQ bisects ∠AOB Now, In ∠BOC and ∠AOC OA = OB [Radius] OC = OC [Common side] AC = BC [Given] Then, by SSS condition: ΔAOC ≅ ΔBOC So, ∠AOC = ∠BOC [By c.p.c.t.] Therefore, PQ bisects ∠AOB. Hence proved.Read more on Sarthaks.com - https://www.sarthaks.com/612362/prove-diameter-circle-which-bisects-circle-bisects-angle-subtended-chord-centre-circle
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