Explicație pas cu pas:
ΔABC triunghi oarecare, AB = 8 cm, AC = 12 cm, m(∢BAC) = 60°, AD bisectoarea unghiului ∢BAC, D ∈ BC, BF ⊥ AD, F ∈ AD, BF ∩ AC = {E}
∢EAF ≡ ΔBAF = 30°
=> ΔEAF ≡ ΔBAF (cazul U.I.)
=> AE ≡ AB = 8 cm
CE = AC - AE = 12 - 8 = 4 cm
ducem BM ⊥ AC, M ∈ AC
[tex]\frac{Aria_{\triangle BCE}}{Aria_{\triangle ABE}} = \frac{ \frac{BM \times CE}{2} }{ \frac{BM \times AE}{2} } = \frac{CE}{AE} = \frac{4}{8} = \frac{1}{2} \\ [/tex]
[tex]\implies \frac{Aria_{\triangle BCE}}{Aria_{\triangle ABE}} = \frac{1}{2} \\ [/tex]
sau:
se observă că ΔABE este echilateral (AE ≡ AB și m(∢BAC) = 60°)
[tex]Aria_{\triangle ABE} = \frac{ {AB}^{2} \sqrt{3} }{4} = \frac{ {8}^{2} \sqrt{3} }{4} = 16 \sqrt{3} \: {cm}^{2} \\ [/tex]
[tex]Aria_{\triangle ABC} = \frac{AB \cdot AC\cdot \sin(\angle BAC) }{2}\\ = \frac{8 \cdot 12 \cdot \sqrt{3} }{2 \cdot 2} = 24 \sqrt{3} \: {cm}^{2} [/tex]
[tex]Aria_{\triangle BCE} = Aria_{\triangle ABC} - Aria_{\triangle ABE} \\ = 24 \sqrt{3} - 16 \sqrt{3} = 8 \sqrt{3} \: {cm}^{2}[/tex]
[tex]\implies \frac{Aria_{\triangle BCE}}{Aria_{\triangle ABE}} = \frac{8 \sqrt{3} }{16 \sqrt{3} } = \frac{1}{2} \\ [/tex]
