Parallelism of two lines in a triangle

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.

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

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 $$FI\parallel BC$$.

• Thank you for sharing your opinion. Solved it now. :) Apr 27, 2021 at 14:51
• Oh, that's a nice observation! Apr 27, 2021 at 17:34
• @Calvin Lin Thanks! Apr 27, 2021 at 17:50

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

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

• Thank you for sharing your opinion. Solved it now. :) Apr 27, 2021 at 14:51