Incircle problem

Question

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.

Answer 1

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 $DB = FC$