AMC 8 2019 (Problem 24)In triangle ABCABCABC, point DDD divides side AC‾\overline{AC}AC so that AD:DC=1:2AD:DC=1:2AD:DC=1:2. Let EEE be the midpoint of BD‾\overline{BD}BD and let FFF be the point of intersection of line BCBCBC and line AEAEAE. Given that the area of △ABC\triangle ABC△ABC is 360360360, what is the area of △EBF\triangle EBF△EBF?(A) 24\text{(A)}\;24(A)24(B) 30\text{(B)}\;30(B)30(C) 32\text{(C)}\;32(C)32(D) 36\text{(D)}\;36(D)36(E) 40\text{(E)}\;40(E)40Related TopicsToolkit 84 — Ratios of AreasToolkit 85 — Menelaus' TheoremHints (9)Hint 1By Toolkit 85 — Menelaus' Theorem, find BFFC\frac{BF}{FC}FCBF.Hint 2Consider △BDC\triangle BDC△BDC and the three collinear points AAA, EEE, and FFF.ADAC⋅FCBF⋅BEED=1\frac{AD}{AC}\cdot\frac{FC}{BF}\cdot\frac{BE}{ED}=1ACAD⋅BFFC⋅EDBE=113⋅FCBF⋅11=1\frac13\cdot\frac{FC}{BF}\cdot\frac11=131⋅BFFC⋅11=1FC=3BFFC=3BFFC=3BFHint 3Assume [AED]=w[AED]=w[AED]=w. Find the areas of the other triangles in terms of www.Hint 4 Hint 5 Hint 6[BDC][ABD]=CDAD=2xx=2\frac{[BDC]}{[ABD]}=\frac{CD}{AD}=\frac{2x}{x}=2[ABD][BDC]=ADCD=x2x=2⟹[BDC]2w=2\Longrightarrow \frac{[BDC]}{2w}=2⟹2w[BDC]=2⟹[BDC]=4w\Longrightarrow [BDC]=4w⟹[BDC]=4wHint 7[BEC]=[BDC]−[EDC]=4w−2w=2w[BEC]=[BDC]-[EDC]=4w-2w=2w[BEC]=[BDC]−[EDC]=4w−2w=2wHint 8[BEF][BEC]=BFBC=z4z=14\frac{[BEF]}{[BEC]}=\frac{BF}{BC}=\frac{z}{4z}=\frac14[BEC][BEF]=BCBF=4zz=41⟹[BEF]=2w4=w2\Longrightarrow [BEF]=\frac{2w}{4}=\frac{w}{2}⟹[BEF]=42w=2wHint 9[ABC]=360[ABC]=360[ABC]=3606w=360⟹w=606w=360\Longrightarrow w=606w=360⟹w=60[EBF]=w2=30[EBF]=\frac{w}{2}=30[EBF]=2w=30Final Answer(B) 303030
