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2007 AMC 12A Problem 6

Problem 6 of 25EasierGeometry

Triangles ABCABC and ADCADC are isosceles with AB=BCAB=BC and AD=DC.AD=DC. Point DD is inside △ABC,\triangle ABC, ∠ABC=40∘,\angle ABC=40^\circ, and ∠ADC=140∘.\angle ADC=140^\circ. What is the degree measure of ∠BAD?\angle BAD?

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Solution

Since △ABC\triangle ABC is isosceles, ∠BAC=12(180∘−∠ABC)\angle BAC=\tfrac12(180^\circ-\angle ABC) =70∘.=70^\circ. Since △ADC\triangle ADC is isosceles, ∠DAC=12(180∘−∠ADC)\angle DAC=\tfrac12(180^\circ-\angle ADC) =20∘.=20^\circ. Therefore ∠BAD=∠BAC−∠DAC\angle BAD=\angle BAC-\angle DAC =70∘−20∘=70^\circ-20^\circ =50∘.=50^\circ. Thus, the correct answer is D.

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Concepts: isosceles triangle · angle chasing

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