Half Of One Side, Third Of Another

• Method 1:

With $D$ being the origin of a grid with standard orientation, draw a line segment from $F$ to $P(2,2)$ . Then as $PF$ has slope $\frac{1}{2}$ and $AP$ has slope $-2$ we see that $\angle APF = 90^{\circ}$ . Also, as $|AP| = |PF| = 2\sqrt{5}$ we see that $\Delta APF$ is a right isosceles triangle at $P$ , and thus that $\angle PAF = \boxed{45^{\circ}}$ .

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• Method 2:

$\angle EAF = 90^{\circ} - (\angle DAE + \angle FAB) = 90^{\circ} - (\tan^{-1}(\frac{1}{2}) + \tan^{-1}(\frac{1}{3})) =$

$90^{\circ} - \tan^{-1}\left(\dfrac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}*\frac{1}{3}}\right) = 90^{\circ} - \tan^{-1}(1) = \boxed{45^{\circ}}$ ,

where the formula $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\dfrac{x + y}{1 - xy}\right)$ was used.

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• Method 3:

Using the cosine rule and Pythagoras, we have that

$|EF|^{2} = |AE|^{2} + |AF|^{2} - 2|AE||AF|\cos(\angle EAF) \Longrightarrow$

$25 = 45 + 40 - 2\sqrt{45}\sqrt{40}\cos(\angle EAF) \Longrightarrow$

$\cos(\angle EAF) = \dfrac{60}{2\sqrt{45*40}} = \dfrac{30}{\sqrt{9*5*4*2*5}} = \dfrac{30}{30\sqrt{2}} = \dfrac{1}{\sqrt{2}}$

$\Longrightarrow \angle EAF = \cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \boxed{45^{\circ}}$ .

Draw a triangle ADF' congruent to triangle ABF. Then angle BAF = angle F'AD. We see that GH = 1 whole 2/3 and HF = 5. Now using Pythagoras theorem we can easily find out AG = 2 sqrt(10) / 3 and AF = 2 sqrt 10. We see that GH / HF = AG / AF. Therefore AH is the angle bisector of angle FAF' by the converse of angle bisector theorem. .'. Required green angle in figure is = Angle 90 / 2 = 45.