1. Olympaid geometry

Consider five points $A, B, C, D$ and $E$ such that $ABCD$ is a parallelogram and
$BCED$ is a cyclic quadrilateral. Let $l$ be a line passing through $A$. Suppose
that $l$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF = EG = EC$.

Prove that $l$ is the bisector of angle $DAB$.

2. Originally Posted by mathwizard
Consider five points $A, B, C, D$ and $E$ such that $ABCD$ is a parallelogram and
$BCED$ is a cyclic quadrilateral. Let $l$ be a line passing through $A$. Suppose
that $l$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF = EG = EC$.

Prove that $l$ is the bisector of angle $DAB$.
I drew a rough sketch of the figure on paper. (I do't know how to draw/graph in computers. Knowing how to do that will surely help a lot in solving geometry problems, for those drawings are exact.)

So triangle FCG is an equilateral triangle because EF = EC = EG.

FA is parallel to FG,
FD is parallel to FC,
Therefore, triangles FAD and FCG are similar....and so are equilateral both.

If a line is drawn through point F, parallel to AD or BC, this line will interesect AB at, say, point H.

AF is a segment of the $l$.
AF is a diagonal of rhombus ADFH.
AF bisects angle DAB then.
Therefore, $l$ bisects angle DAB also ----proven.

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If by not through the characteristics of a rhombus, AF can be proven as bisecting angle DAB by considering both triangles DAF and HAF are equilateral (and equiangular) triangles. That angle DAF = angle HAF = 60 degrees. Etc.

3. Originally Posted by ticbol
I drew a rough sketch of the figure on paper. (I do't know how to draw/graph in computers. Knowing how to do that will surely help a lot in solving geometry problems, for those drawings are exact.)

So triangle FCG is an equilateral triangle because EF = EC = EG.
Hmm I did not understand this part . Can you explain?

4. Originally Posted by Isomorphism
Hmm I did not understand this part . Can you explain?
Yes, proving the triangle FCG an equilateral is the weakest part of my proof.
I saw it the first time I chanced upon the problem but I did not make a reply then because it was hard to prove triangle FCG is equilateral.

The second time, I did post the reply above because I was banking on my sketch that triangle FCG is equilateral. If so, then the 3 radii EF, EC and EG divide triangle FCG into 3 congruent isosceles triangles, making FCG an equilateral triangle.
That was an assumption on my part, FCG being a 60-60-60 triangle based on the sketch only.

Maybe the quadrilateral BCED being cyclic has something to do with the sketch showing FCG being equilateral. I have not investigated that yet.

For now, I let the proof hung. I'd go more into it when I have time.

5. Originally Posted by ticbol
Yes, proving the triangle FCG an equilateral is the weakest part of my proof.

For now, I let the proof hung. I'd go more into it when I have time.
After carefull sketching by hand, I found out that cyclic quadrilateral BCED doesn't help. Other than it looks like an isosceles trapezoid, BCED does not lead to FCG being equilateral.

So the next approach is to prove that angle FCE is 30 degrees.
In isosceles triangle FEC, if angle FCE, and so angle CFE too, were 30 degrees each, then the apex angle, or angle FEC, should be 120 degees.

In triangle FCG, if side FC is horizontal:
Point E is the centroid of triangle.
Extend radius EG vertically upwards to point, say J, on side FC.
Extension EJ is along the perpendicular bisector of FC.
Since point E is the centroid of triangle FCG, and therefore E is 1/3 of JG from J, then EG = 2(EJ)
Since EC = EG = 2(EJ), then angle JCE = 30 degrees....in a right triangle, if hyp is twice the opp, then the angle is 30 degrees.

Hence isosceles triangle FEC is a 30-120-30 degree triangle.
Blah, blah, so are isosceles triangles CEG and GEF.
Blah, blah, hence triangle FCG is an equilateral triangle.

Tie that up with my first post, therefore, blah, blah.....