Ceva's theorem applied on a triangle but I am stuck on a step.

Question

Q. Let $ABC$ be a triangle. Let $r$ and $s$ be the angle bisectors of angle $ABC$ and angle $BCA$, respectively. The points $E$ in $r$ and $D$ in $s$ are such that $AD || BE$ and $AE || CD$. The lines $BD$ and $CE$ cut each other at $F$. The point $I$ is the incenter of $ABC$. Show that if $A, F$ and $I$ are collinear, then $AB = AC$.
Sol. Let $AI$ meet $BC$ at $K$. Apply Ceva’s theorem to $\triangle FBC$, where the cevians $FK, BE, CD$ meet at $I$. Then,we get $$\frac{BK}{KC}\cdot\frac{CE}{EF}\cdot\frac {FD}{DB} =1$$ Since $\frac{FD}{DB}=\frac{FA}{AI}$ and $\frac{CE}{EF}=\frac{AI}{FA}$, then we see immediately that $BK = KC$, which implies that the angle bisector of $\triangle ABC$ is its median, which shows that $AB = AC$.

My geometry is real bad and I need help in understanding why the statement "Since $\frac{FD}{DB}=\frac{FA}{AI}$ and $\frac{CE}{EF}=\frac{AI}{FA}$" is true, as it somehow isn't obvious to me. Thanks!

P.S. Sorry in advance if my question is too easy and I look hopeless.

Answer 1

The incenter $I$ of $\triangle ABC$ lies on all three angle bisectors and hence lies on $CD$ and $BE$.

We are given that $AD||BE$ and $AE||CD$. Hence $AD||BI$ and $AE||CI$.

From the above, and under the given assumption that $A$, $F$ and $I$ are collinear:

• $\triangle ADF$ and $\triangle IBF$ are similar triangles, so $\frac{FD}{DB}=\frac{FA}{AI}$ holds.
• $\triangle AEF$ and $\triangle ICF$ are similar triangles, so $\frac{CE}{EF}=\frac{AI}{FA}$ holds.