If Chord AB Of A Circle Is Parallel To The tangent at C,Then prove that Ac=Bc
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AB is a chord of the circle, and by taking a point C on the circle ,A tangent is drawn such that ,
→AB ║Line l(Tangent at point C)
Join AC and BC.
Find center of circle O , by drawing perpendicular bisectors of sides AC and CB.
Draw , OM ⊥ AB, and Join O and C.
As, ∠OCN= 90°, and ∠OCS=90°→→Angle between line joining from center to the point of contact on the circle from where the tangent passes is of 90°.
Also, AM = MB →→[⊥ from the center to the chord bisects the chord]----(1)
∠BMC +∠NCO=180°
So, Points,M , O and C lie in a line.
In Δ C MA and Δ CM B
Side CM is common.
∠ C MA = ∠CM B=90°
AM= MB---[Proved at 1]
→ Δ C MA ≅ Δ CM B ⇒⇒[S AS]
→AC= BC→→[C PCT]
Answer:8
Step-by-step explanation:
Secondary SchoolMath 5+3 pts
If Chord AB Of A Circle Is Parallel To The tangent at C,Then prove that Ac=Bc
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anildeshmukh
Anildeshmukh · Ace
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CarlynBronk Ambitious
AB is a chord of the circle, and by taking a point C on the circle ,A tangent is drawn such that ,
→AB ║Line l(Tangent at point C)
Join AC and BC.
Find center of circle O , by drawing perpendicular bisectors of sides AC and CB.
Draw , OM ⊥ AB, and Join O and C.
As, ∠OCN= 90°, and ∠OCS=90°→→Angle between line joining from center to the point of contact on the circle from where the tangent passes is of 90°.
Also, AM = MB →→[⊥ from the center to the chord bisects the chord]----(1)
∠BMC +∠NCO=180°
So, Points,M , O and C lie in a line.
In Δ C MA and Δ CM B
Side CM is common.
∠ C MA = ∠CM B=90°
AM= MB---[Proved at 1]
→ Δ C MA ≅ Δ CM B ⇒⇒[S AS]
→AC= BC→→[C PCT]