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In triangle $ABC$, point $M$ is the midpoint of $BC$ and $N$ is on the angle bisector of $\angle A$ such that $MN \parallel AB$. Prove that $MN = \dfrac{|b − c|}{2}$.

Attempt: I drew it out and noticed that $\triangle{DMN}\sim\triangle{DBA}$. I am not sure how to use the fact that $M$ is the midpoint yet, but I think the angle bisector theorem will help. We have $\dfrac{BD}{c} = \dfrac{a/2+DM}{b}$. Then by the similarity $\dfrac{DM}{BD} = \dfrac{MN}{c}$. I am not sure what to do next.

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  • $\begingroup$ Note that by angle bisector theorem, we can easily take out values of $BD$ and $CD$. Putting them in the similarity condition will give you the answer. $\endgroup$ – Sawarnik Feb 5 '16 at 18:06
  • $\begingroup$ There's another alternative way: let $E$ be the intersection of $MN$ and $AC$. Then, as $\angle NAE = \angle ENA$, we have $NE=AE$. Thus, $$MN=NE-ME=AE-ME=\frac{b-c}2$$. $\endgroup$ – Sawarnik Feb 5 '16 at 18:11
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You are on the right track. Let $BD=x,DM=y$. Then, $$\frac{x}{c}=\frac{y+a/2}{b}$$ The other equality in $x$ and $y$ come from the fact that $$x+y=\frac{a}{2}$$ This is where the midpoint comes in. Now solve for $x$ and $y$, and use the condition of similarity $$\frac{c}{x}=\frac{MN}{y}$$ This gives $$MN=\frac{b-c}{2}$$ EDITS: There is an alternative case in which having the same definitions would give $$x-y=\frac{a}{2}$$ $$\frac{x}{c}=\frac{a/2-y}{b}$$ Solving these gives the other part and using the similarity condition (you can check this still holds), $$MN=\frac{c-b}{2}$$ Hope it helps !

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  • $\begingroup$ The problem with this solution is that $x+y$ is not always equal to $a/2$. $\endgroup$ – user19405892 Feb 5 '16 at 18:20
  • $\begingroup$ You are right, thanks for pointing that out. Fixing. $\endgroup$ – Ashish Gaurav Feb 5 '16 at 18:23

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