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The question is that:

Let $ABCD$ be a parallelogram. Let $X$ and $Y$ be points on $AB$ and $BC$ respectively, such that $AX=CY$. Prove that the intersection of lines $AY$ and $CX$ lies on the angle bisector of $\angle ADC$.

I have put the parallelogram on coordinate plane and proved that. However, are there any way such that it can be solved algebraically?

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  • $\begingroup$ Think in terms of vectors. $\endgroup$ – Zain Patel Jul 3 '16 at 14:42
  • $\begingroup$ The answer below shows an approach through Ceva's theorem and the bisector theorem. Truth to be told, it is equivalent to using trilinear or barycentric coordinates, not so different from using cartesian coordinates or vectors, as Zain Patel suggested. $\endgroup$ – Jack D'Aurizio Jul 3 '16 at 16:10
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Let $P=AY\cap CX$, $\,A'=DA\cap CX$, $\,C'=DC\cap AY$, $\,AB=x$, $\,BC=y$, $\,AX=CY=z$.

Since $A'AX$ and $A'DC$ are similar triangles, $\frac{A'A}{A'D}=\frac{z}{x}$.
Since $C'CY$ and $C'DA$ are similar triangles, $\frac{C'C}{C'D}=\frac{z}{y}$.

Let $Q=DP\cap A'C'$. By Ceva's theorem, $$ \frac{A'Q}{QC'}=\frac{CD}{C'C}\cdot \frac{AA'}{DA} = \frac{y-z}{z}\cdot\frac{z}{x-z} = \frac{y-z}{x-z}.$$ By the angle bisector theorem, $Q$ lies on the angle bisector of $A'DC'$ iff $\frac{A'Q}{QC'}=\frac{A'D}{DC'}$.
Given the previous similarities, that is straightforward to check: $$ A'D = \frac{xy}{x-z},\qquad C'D = \frac{xy}{y-z}.$$

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