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Question
The given figure shows a circle with centre O and BCD is tangent to it at C. Show that : ∠ACD + ∠BAC = 90°.
Solution
Join OC.
BCD is the tangent and OC is the radius.
∴ OC ⊥ BD
`=>` ∠OCD = 90°
`=>` ∠OCA + ∠ACD = 90°
But in ΔOCA
OA = OC ...(Radii of same circle)
∴ ∠OCA + ∠OAC
Substituting (i)
∠OAC + ∠ACD = 90°
`=>` ∠BAC + ∠ACD = 90°
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