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Let $ABC$ be an equilateral triangle and let $D, E \in (BC), F\in (AD), G\in (AE)$ such that $m(\angle BAD)=m(\angle DAE)=m(\angle EAC)=m(\angle ABF)=m(\angle CBG)$. The bisector of $\angle ABC$ intersects $(AE)$ in $I$. Prove that:
a) $BEG$ is an isosceles triangle.
b) $AGB$ and $FDB$ are similar triangles.
c) $FI$ and $BC$ are parallel.
I've managed to solve a) and b) and I would say that these should help me to finish c) as well. I would appreciate any tip/help with this final item.

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  • $\begingroup$ What's the context of this problem? EG Can I use Ceva's theorem (in trigonometric form)? $\endgroup$
    – Calvin Lin
    Apr 26, 2021 at 23:17

2 Answers 2

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Observe that $BI$ is the perpendicular bisector of $AC$ and hence $\angle ACI =20^{\circ}$ . Thereafter points $F$ and $I$ are symmetric about the altitude on the side $BC$ from vertex $A$ and thus $FI\parallel BC$.

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    $\begingroup$ Thank you for sharing your opinion. Solved it now. :) $\endgroup$
    – Alchimist
    Apr 27, 2021 at 14:51
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    $\begingroup$ Oh, that's a nice observation! $\endgroup$
    – Calvin Lin
    Apr 27, 2021 at 17:34
  • $\begingroup$ @Calvin Lin Thanks! $\endgroup$
    – Limestone
    Apr 27, 2021 at 17:50
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Note: I didn't use a) and b), and I don't think they are relevant to c).

The difficulty here is that the line $FI$ isn't nicely defined. So, we should try to work without using the line $FI$.

Approach 1:

Hint: Show that $F, I$ are symmetric about the height from A.
Hence, $FI \parallel BC$.

DR SK has an elegant approach.

The most direct way I can think of is $ \angle ACI = 20^\circ$ by Ceva's theorem.

Approach 2:
Hint: Show that the height of $F, I$ from base $BC$ are the same.

The most direct way I can think of is calculating the heights using trigonometry and sine rule

$ \sin 30^\circ \sin 40^\circ = \frac{1}{2} \times 2 \sin 20^\circ \cos 20^\circ = \sin 20^\circ \sin 70^\circ $

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  • $\begingroup$ Thank you for sharing your opinion. Solved it now. :) $\endgroup$
    – Alchimist
    Apr 27, 2021 at 14:51

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