in the figure, O is the centre of the circle, Seg AB is diameter. At the pt C on the circle the tangent CD is drawn. Line BD is also a tangent to the circle at pt B .Show that seg OD || chord AC
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24
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riri01:
how can we prove through showing congruent angles? like with showing the similarity of triangles...
if we show two triangles congruent,we get corresponding of triangles which are not proven.with that we can prove that is a parallelogram
yess...but is there any alternate method? like without proving parallelogram?
i dont no.i studied this method only.
where r u from?
ok...anyway thanks for the answer!
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15
seg OD || chord AC
Step-by-step explanation:
in Δ OCD & Δ OBD
OC = OB ( radius)
CD = BD ( common tangent)
OD = OD common
=> Δ OCD ≅ Δ OBD
=> ∠COD = ∠BOD
in ΔAOC
OA = OC ( radius)
=> ∠OCA = ∠OAC
∠COB = ∠OCA + ∠OAC ( extriore angle of ΔAOC)
=> ∠COD + ∠BOD = ∠OCA + ∠OCA
=> ∠COD + ∠COD = 2∠OCA
=> 2∠COD = 2∠OCA
=> ∠COD = ∠OCA
=> AC ║ OD ( alternate angle)
QED
Proved
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