This is sort of an involved question. I've proved parts of it already but now I'm stuck. Here is the question:

Let ∆ABC be such that AB is not congruent to AC. Let D be the point of intersection of the bisector of $\angle A$ and the perpendicular bisector of side BC. Let E, F, and G be the feet of the perpendiculars dropped from D to lines AB, AC, BC ⃡, respectively. Prove that

(a) D lies outside of the triangle on the circle through ABC [which I did by proving that both the bisector of $\angle A$ and the perpendicular bisector of segment BC bisect arc BC on the circumcircle of $\triangle ABC$.]

And given that one of E or F lies inside the triangle and the other outside, prove

(b) E, F, and G are collinear.

So my ideas so far are:

-my strategy is to prove that EF intersects the midpoint of BC (point G).

-By the crossbar theorem, we know that segments EF and BC intersecting AD exist. (although I don't know if it's completely necessary to state it for BC since it's obvious that the bisector of $\angle A$ intersects opposite side BC.)

-Since it is given that "one of E or F lies inside the triangle and the other outside", we know that EF must intersect BC at some point.

-I think that the angle bisector theorem may be helpful (if a point is on the angle bisector then this point is equidistant from the sides of the angle) because from there we can construct congruent triangles and maybe that will be enlightening.

Before I began the problem, I constructed everything with a straight-edge and compass so I have a visual of everything.

I think that $\triangle GFC$ and $\triangle EGB$ may be congruent because $\angle G$ would be the same measurement in both and we know that BG is congruent to CG. I don't know if that would help show that E, G, and F are collinear though.

Also, if anyone knows why these proofs would be useful, I would love to know. I think I read somewhere that it may be useful in navigation at sea or something. That's super interesting if it's true.


Hint: You already showed that $D$ lies on the circumference of $ABC$.

Hence, apply the theorem which says that the foot of the perpendiculars from $D$ are collinear.

If you want to prove that, consider the cyclic quads $DEGC$ and $DFGB$.


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