The given figure shows a circle with center O and BCD is tangent to it at C. Show that : $∠ A C D + ∠ B A C = 90^{\circ}$ .

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Solution

REF.Image
to prove :-
$∠ A C D + ∠ B A C = 90^{\circ}$
construct OC
$∵ B D$ is tangent at point C.
$∴ O C D = 90^{\circ}$
$∠ O C A + ∠ A C D = 90^{\circ}$ - (i)
In $Δ A O C$
$O C = O A = r$
$∴ O C A = O A C$ _ (ii)
$∴$ from (i) & (ii)
$∠ O A C + ∠ A C D = 90^{\circ}$
$∠ B A C + ∠ A C D = 90^{\circ}$

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