Let the quadrilateral be [latex]ABCD[/latex]. Using the provided angles, we can find that [latex]∠BCA=90°[/latex] and therefore point [latex]C[/latex] belongs to the circle with diameter [latex]AB[/latex]. Also, we can see that [latex]△AOD[/latex] is isosceles, [latex]OA=OD=OB[/latex], and point [latex]D[/latex] also lies on the circle with diameter [latex]AB[/latex].
Using the provided angles, we can find that [latex]∠BCA=90°[/latex] and therefore point [latex]C[/latex] belongs to the circle with diameter [latex]AB[/latex]. Also, we can see that [latex]△AOD[/latex] is isosceles, [latex]OA=OD=OB[/latex], and point [latex]D[/latex] also lies on the circle with diameter [latex]AB[/latex].
Therefore, [latex]ABCD[/latex] is inscribed in a circle with center [latex]O[/latex], and [latex]∠ACD=∠AOD/2=40°[/latex].
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