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In triangle ABC, M is the midpoint of BC, AB=12, and AC=16. Points E and F are taken on AC and AB respectively, and EF and AM intersect at G. If AE=2AF then what is EG/GF?

Oct 13, 2019

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In triangle ABC,
M is the midpoint of BC, AB=12, and AC=16.
Points E and F are taken on AC and AB respectively, and EF and AM intersect at G.
If AE=2AF then what is EG/GF?

$$\text{Let \angle EAG=\epsilon_1 } \\ \text{Let \angle AMC=\epsilon_2 } \\ \text{Let \angle AGE=\epsilon_3 } \\ \text{Let \angle GAF=\delta_1 } \\ \text{Let \angle AMB=\delta_2=180^\circ-\epsilon_2  } \\ \text{Let \angle AGF=\delta_3=180^\circ-\epsilon_3  } \\ \text{Let EG=\mathbf{x} } \\ \text{Let GF=\mathbf{y} } \\ \text{Let AF=f } \\ \text{Let AE=2AF=2f }$$

1. sin-rule

$$\begin{array}{|lrcll|} \hline (1) & \dfrac{\sin(\epsilon_1)}{\dfrac{CB}{2}} &=& \dfrac{\sin(\epsilon_2)}{16} \\\\ \hline & \dfrac{\sin(\delta_1)}{\dfrac{CB}{2}} &=& \dfrac{\sin(180^\circ-\epsilon_2)}{12} \\\\ (2) & \dfrac{\sin(\delta_1)}{\dfrac{CB}{2}} &=& \dfrac{\sin(\epsilon_2)}{12} \\\\ \hline & \dfrac{CB}{2}\sin(\epsilon_2) = 16\sin(\epsilon_1)&=& 12\sin(\delta_1) \\ & 16\sin(\epsilon_1)&=& 12\sin(\delta_1) \\\\ & \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} &=& \dfrac{12}{16} \\\\ (3) & \mathbf{ \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} } &=& \mathbf{ \dfrac{3}{4} } \\ \hline \end{array}$$

2. sin-rule

$$\begin{array}{|lrcll|} \hline (4) & \dfrac{\sin(\epsilon_1)}{x} &=& \dfrac{\sin(\epsilon_3)}{2f} \\\\ \hline & \dfrac{\sin(\delta_1)}{y} &=& \dfrac{\sin(180^\circ-\epsilon_3)}{f} \\\\ (5) & \dfrac{\sin(\delta_1)}{y} &=& \dfrac{\sin(\epsilon_3)}{f} \\\\ \hline & \dfrac{\sin(\epsilon_3)}{f} = \dfrac{2\sin(\epsilon_1)}{x} &=& \dfrac{\sin(\delta_1)}{y} \\\\ &\dfrac{x}{y} &=& 2\cdot \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} \quad | \quad \dfrac{\sin(\epsilon_1)}{\sin(\delta_1)} = \dfrac{3}{4} \\\\ &\dfrac{x}{y} &=& 2\cdot \dfrac{3}{4} \\\\ & \mathbf{ \dfrac{x}{y} } &=& \mathbf{ \dfrac{3}{2} } \\ \hline \end{array}$$

$$\mathbf{\dfrac{EG}{GF} = \dfrac32}$$

Oct 15, 2019