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2007 AMC 10A Problem 8

Problem 8 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∘−40∘)=70∘.\angle BAC = \tfrac12(180^\circ - 40^\circ) = 70^\circ. Since △ADC\triangle ADC is isosceles, ∠DAC=12(180∘−140∘)=20∘.\angle DAC = \tfrac12(180^\circ - 140^\circ) = 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

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