In triangle $ABC,$ the points $D$ and $E$ are the intersections of the angular bisectors from $C$ and $B$ with the sides $AB$ and $AC,$ respectively. Points $F$ and $G$ on the extensions of $AB$ and $AC$ beyond $B$ and $C,$ respectively, satisfy $BF = CG = BC.$ Prove that $F G \parallel DE.$
Problem
Source: Baltic Way 2016, Problem 16
Tags: geometry
Tintarn
06.11.2016 02:05
Just note that
\[\frac{AD}{AE}=\frac{\frac{bc}{a+b}}{\frac{bc}{a+c}}=\frac{a+c}{a+b}=\frac{AF}{AG}\]and the conclusion follows immediately.
Kezer
06.11.2016 02:19
A longer solution, if you fail to notice the explicit length of $|AD|$ or $|AE|$ as in Tintarn's solution: Let $|AD|=d,|DB|=e,|AE|=f,|EC|=g$ and $|BC|=a$. Then \[ \frac{d}{e}=\frac{f+g}{a} \quad \text{and} \quad \frac{f}{g}=\frac{d+e}{a}. \]It suffices to prove \[ \frac{d}{f} = \frac{d+e+a}{f+g+a} \iff dg+da=ef+af \iff dg+eg=af \iff \frac{f}{g}=\frac{d+e}{a}. \]Done. $\hfill \square$
AlastorMoody
30.06.2019 12:21
This is just Pappus' Theorem on $EBGFCD$