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2015 AMC 12B Problem 13

Problem 13 of 25IntermediateGeometry

Quadrilateral ABCDABCD is inscribed in a circle with ∠BAC=70∘,\angle BAC = 70^\circ, ∠ADB=40∘,\angle ADB = 40^\circ, AD=4,AD = 4, and BC=6.BC = 6. What is AC?AC?

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Solution

Angles BACBAC and BDCBDC subtend arc BC,BC, so ∠BDC=70∘.\angle BDC = 70^\circ. Then ∠ADC=∠ADB\angle ADC = \angle ADB +∠BDC=110∘.+ \angle BDC = 110^\circ. Since ABCDABCD is cyclic, ∠ABC=180∘−110∘=70∘\angle ABC = 180^\circ - 110^\circ = 70^\circ =∠BAC.= \angle BAC. Thus △ABC\triangle ABC is isosceles with AC=BC=6.AC = BC = 6. Thus, the correct answer is B.

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Concepts: cyclic quadrilateral · inscribed angle · isosceles triangle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.