I found a rather interesting exercise which I have some problems with.
A line parallel with $AB$ and the triangle $\Delta ABC $ intercepts the sides $AC$ and $BC$ at the points $A_1$, $B_1$.
Show that the lines $A_1B$, $AB_1$ and the median from $C$ is concurrent. In other words, show that the lines intersect at a single point.
Line of thought: Apply Cevas Theorem
If $A1,B1,C1$ are points on $AC$, $BC$ and $AB$ such that \begin{equation}\frac{A C_1}{C_1B}\cdot \frac{B B_1}{B_1 C} \cdot \frac{CA_1}{A_1A}=1 \quad (1)\end{equation} then $AB_1$, $BA_1$ and $CC_1$ are concurrent.
The median from $C$ divides the side $AB$ in half such that $AC_1=C_1B$. Therefore $\frac{AC_1}{C_1B}=1$.
$(1)$ now becomes \begin{equation} \frac{BB_1}{B_1C}\cdot \frac{CA_1}{A_1A}=1 \quad (2) \end{equation}
Maybe we could use Thael's theorem triangle tells us that the ratio of any two corresponding sides of an equiangular triangle is always irrespective of their sizes. Thaels would give us that $$\frac{BB_1}{B_1C}=\frac{A_1A}{CA_1}$$ and if we insert this in $(2)$ the equality is correct.
This is my attempt but I'm pretty sure this is wrong. Seems like I'm doing some circular reasoning in the end. Any suggestions/help or even a completely other solution would be greatly appreciated!
Best Answer
Let $AB_1\cap BA_1=\{D\}$, $CD\cap A_1B_1=\{E\}$ and $CD\cap AB=\{F\},$
$A_1E=a$, $EB_1=b$, $AF=c$ and $FB=d$.
Thus, $$\frac{a}{c}=\frac{CE}{CF}=\frac{b}{d}$$ and $$\frac{a}{d}=\frac{ED}{DF}=\frac{b}{c},$$ which gives $a=b$, $c=d$ and we are done!