Incircle problem Triangle ABC has incircle $ \beta$ which meets BC at D. A diameter of the incircle has endpoints E and D. A line joining A and E meets BC at F. Given that $DC \gt BC$. Prove that $BD =FC$
I couldn't find any synthetic geometry methods, so I resorted to coordinate geometry. I took the incircle as a circle with radius 1 and centre (0,1). $B \equiv (-x,0) , C \equiv (y,0)$
I found $A \equiv (\frac{x-y}{xy-1} , \frac{2xy}{xy-1} )$
Then on extending AE we get $ F \equiv (y-x,0)$ and hence it is proved that $\overline{BD} = \overline{FC}$. 
I hope anyone could provide me proof with Euclidean geometry, which is more intuitive and brainy than bashing.
Note: Please help in putting a suitable title.
 A: Let $k$ be the incircle of triangle $ABC$. Draw a line through point $E$ tangent to circle $k$ and let $B'$ be its intersection point with edge $AB$ and $C'$ be its intersection point with edge $BC$. Then $B'C'$ is parallel to edge $BC$ because as tangents to circle $k$ both $B'C'$ and $BC$ are orthogonal to diameter $DE$. Furthermore, the triangle $AB'C'$ is similar to triangle $ABC$.

Hence, a homothety with respect to point $A$ sending point $E$ to point $F$ maps triangle $AB'C'$ to the larger triangle $ABC$. Then the incircle of triangle $ABC$, which is an excircle (escribed circle) of triangle $AB'C'$, is sent by this homothety to the excircle $k_{BC}$ (escribed circle) on the side of edge $BC$. Then the excircle $k_{BC}$ touches the edge $BC$ at point $F$, because its preimage $k$ touches $B'C'$ at point $E$. Therefore $FC = T'C = p - AC$, where $p$ is the half perimeter of triangle $ABC$. Since  incircle $k$ touches edge $BC$ at point $D$, we have $DB = SB = p - AC$. Hence $DB = FC$     
A: 
Draw line parallel to $BC$ through $E$, intersecting $AB$ at $M$ and $AC$ at $N$. $BCNM$ is a tangential quadrilateral, and even better: it is a trapezoid! This lead to a very special properties: $BD\cdot ME=NE\cdot CD=r^{2}$
$$
\begin{aligned}
BD\cdot (ME+NE)&=NE\cdot (BD+CD)\\
BD\cdot MN&= NE\cdot BC\\
BD&=\frac{NE}{MN}\cdot BC\\
BD&=CF
\end{aligned}
$$
note: if You are wondering why tangential trapezoid have this special properties, try to tinker with the fact that $\angle BOM=\angle CON=\frac{\pi}{2}$ with $O$ the incircle's center
A: First, we focus on the segment $FC$. Since $B'C'\, || \, BC$, by Thales' intercept theorem, the triangles $\Delta\, AB'C'$ and $\Delta\, ABC$ are similar and therefore
$$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = m$$
By the same fact, that $EC'\, || \, FC$, and the same theorem:
$$\frac{FC}{EC'} = \frac{AC}{AC'} = m$$
so $FC = m\, EC'$.
By the tangency of the incircle $k$ of triangle $\Delta \, ABC$ to the (extended) sides of triangle $\Delta\, AB'C'$,
$$EC' = TC' = TA - AC' = p' - AC'$$ where $p'$ is half of the perimeter of $\Delta\, AB'C'$. This implies
$$FC = m\, EC' = m\, (p' - AC') = m\,p' - m\, AC' = p - AC$$ where $p$ is  half of the perimeter of $\Delta\, ABC$.
Now we focus on the segment $BD$. Since $k$ is the incircle of $\Delta \, ABC$, 
$$BD = p - AC$$ Therefore
$$FC = p - AC = BD$$ 
