+5 In triangle ABC, angle BAC measures 20° and angle BCA measures 23°. Points D and E are inside triangle ABC such that angles BAD, DAE, and EAC have the same measure and angles BCD, DCE, and ECA have the same measure. Lines DE↔ and AC↔ intersect at F. Compute the degree measure of angle AFD.
+15001 Hi aw007!
\(Point\ A\ is\ the\ origin.\ \overline{AB}=1\\\)
\( \color{blue}Point\ C:\\f_{AC}(x)=tan 20°\cdot x\\ f_{BC}(x)=tan 43°\cdot (x-1)\\ tan 20°\cdot x=tan 43°\cdot (x-1)\\ (tan 43°-tan 20°)\cdot x-tan 43°=0\\ {\color{blue}x_C}=\frac{tan 43°}{tan 43°-tan 20°}=\color{blue}1.64018\\ {\color{blue}y_C}=\frac{tan 43°\cdot tan 20°}{tan 43°-tan 20°}=\color{blue}0.59698\)
\(\color{blue}Point\ D:\\m_{AD}=tan\ \frac{20°}{3}\\ f_{AD}(x)=tan\ \frac{20^\circ}{3}\cdot x\\ y_D=tan\ \frac{20^\circ}{3}\cdot x_D\\ f_{CD}(x)=m(x-x_C)+y_C\\ m_{CD}=tan\ (180°-(180-20-23)°-\frac{23°}{3})=tan\ 35.\overline 3°\\ y_D=m_{CD}(x_D-x_C)+y_C=tan\ 35.\overline 3(x_D-1.64018)+0.59698\)
\(tan\ \frac{20^\circ}{3}\cdot x_D=tan\ 35.\overline 3°\cdot x_D-tan\ 35.\overline 3°\cdot 1.64018+0.59698\\ x_D=\frac{tan\ 35.\overline 3°\cdot 1.64018-0.59698}{tan\ 35.3\overline 3°-tan\ \frac{20°}{3}}\\ \color{blue}x_D=0.95564\\ y_D=tan\ \frac{20^\circ}{3}\cdot x_D\\ \color{blue}y_D=0.11170\)
\(\color{blue}Point\ E:\\ f_{AE}(x)=tan\ \frac{2\cdot 20^\circ}{3}\cdot x\\ y_E=tan\ \frac{40^\circ}{3}\cdot x_E\\ f_{CE}(x)=m(x-x_C)+y_C\\ m_{CE}=tan\ (180°-(180-20-23)°-\frac{2\cdot 23°}{3})=tan\ 27.6\overline 6°\\ y_E=m_{CE}(x_E-x_C)+y_C=tan\ 27.6\overline 6(x_E-1.64018)+0.59698\)
\(tan\ \frac{40^\circ}{3}\cdot x_E=tan\ 27.6\overline 6(x_E-1.64018)+0.59698\\ tan\ \frac{40^\circ}{3}\cdot x_E=tan\ 27.6\overline 6\cdot x_E-tan\ 27.6\overline 6\cdot 1.64018+0.59698\\ .\\ x_E=\frac{tan\ 27.6\overline 6\cdot 1.64018-0.59698}{tan\ 27.6\overline 6-tan\ 13.3\overline3°}\\ \color{blue}x_E=0.91524\)
\(y_E=tan\ \frac{40^\circ}{3}\cdot x_E=tan\ \frac{40^\circ}{3}\cdot 0.91524\\ \color{blue}y_E=0.21692\)
\(m_{DE}=\frac{y_E-y_D}{x_E-x_D}=\frac{0.21692-0.11170}{0.91524-0.95564}\\ m_{DE}=-2.60446\)
\(\angle ADF=\angle DAB+atan \ m_{DE}=\frac{20°}{3}-atan\ (-2.60446)\\ \angle ADF=75\frac{2}{3}^\circ\\ \color{blue}The\ degree\ measure\ of\ angle\ AFD\ is\ 75\frac{2}{3}^\circ.\)
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