Connected Triangles' Angle

Finding the length of $\overline{CD}$ and the measurement of $\angle BDC$ :

Triangle $ABC$ has a shown width of $10$ . It would have a height of $2\sqrt{14}$ . Triangle $AHE$ has a width of $16$ , which would have a height of $\sqrt{17}$ . Subtracting the height of $\triangle AHE$ from the height of $\triangle ABC$ would give us $\approx 3.36$ which is equal to the length of $CD$ . $$ We can now compute for $\angle BDC$ . $\small \tan\angle BDC = \frac{10}{3.36} \implies \angle BDC \approx 71.43^{\circ}$

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Finding the length of $\overline{EF}$ and the measurement of $\angle FDE$ :

Line segment $\overline{AJ}$ is equal to the length of half of $\overline{BC}$ plus the length of $\overline{DE}$ . Line segment $\overline{JK}$ would then have a length of $1$ . Using the Intersecting Chords Theorem ( $\overline{LJ} × \overline{JK} = \overline{EJ} × \overline{FJ}$ ), we get that the length of $\overline{EF}$ is $2\sqrt{17}$ . $$ We can now compute for $\angle FDE$ . $\small \tan \angle FDE = \frac{2\sqrt{17}}{3} \implies \angle FDE \approx 70.01^{\circ}$

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Having the measurements of the necessary angles, we can calculate for the answer to the question.

$\scriptsize \begin{aligned} \angle BDC + \angle CDE + \angle FDE + \angle BDF = & \space 360^{\circ} \\ 71.43^{\circ} + 90^{\circ} + 70.01^{\circ} + \angle BDF = & \space 360^{\circ} \\ 231.44^{\circ} + \angle BDF = & \space 360^{\circ} \\ \angle BDF = & \space 360^{\circ} - 231.44^{\circ} \\ \angle BDF = & \space \boxed{128.56^{\circ}}\end{aligned}$