.
-
numbers 
each equal to either 
or 
.
so that, for any 
and indices 
so that 
for all 
,
and 
such that
Y by Adventure10, Mango247
Let 
be a triangle. Points 
and 
lie on sides 
and 
, respectively, such that 
. Segments 
and 
meet at 
. The bisector of angle 
meet segments and at 
and 
, respectively. Prove that 
.
be a triangle. Points 
and 
lie on sides 
and 
, respectively, such that 
. Segments 
and 
meet at 
. The bisector of angle 
meet segments and at 
and 
, respectively. Prove that 
.
are the 
and 
.![\[\frac{QD}{QA}=\frac{CD}{AC} \implies \frac{AD}{QA}=\frac{CD+AC}{AC}=\frac{CD+CE+AE}{AC}.\]](http://latex.artofproblemsolving.com/f/5/c/f5c19abc34178a2dcace41939ab72a769a95aa7d.png)
![\[\frac{EC}{ER}=\frac{BD+EC+CD}{BE}\]](http://latex.artofproblemsolving.com/9/4/f/94ffeba8d2a23e2f2f0915f061cefa410083844e.png)
.Cancelling two common terms from these two equations we get, ![\[\frac{QA}{AC}\cdot \frac{EC}{ER}=\frac{AD}{BE}\]](http://latex.artofproblemsolving.com/0/8/c/08cdc68f9007dffc0c29d7c7737644f857470657.png)
with the transversal 
we get,![\[ \frac{ER}{PR}\cdot \frac{AC}{EC}\cdot \frac{PQ}{QA}= 1 \implies \frac{PQ}{AD}=\frac{PR}{BE}\]](http://latex.artofproblemsolving.com/1/7/6/1766bc54c3dec573f12fd027e9bdafaa97efcd81.png)
And since 
we have:
Now using Menelaus Theorem in 
wrt 
