jred wrote: In $\triangle$ABC, points D, M lie on side BC and AB respectively, point P lies on segment AD. Line DM intersects segments BP, AC (extended part), PC (extended part) at E, F and N respectively. Show that if DE=DF, then DM=DM. What do you men prove that $DM=DM$ ??? Do you mean $DM=DN$ ???

War-Hammer wrote: jred wrote: In $\triangle$ABC, points D, M lie on side BC and AB respectively, point P lies on segment AD. Line DM intersects segments BP, AC (extended part), PC (extended part) at E, F and N respectively. Show that if DE=DF, then DM=DM. What do you men prove that $DM=DM$ ??? Do you mean $DM=DN$ ??? yes, it should be DM=DN. i just made a silly mistake. i really appreciate your questioning.

jred wrote: In $\triangle$ABC, points D, M lie on side BC and AB respectively, point P lies on segment AD. Line DM intersects segments BP, AC (extended part), PC (extended part) at E, F and N respectively. Show that if DE=DF, then DM=DN. It is butterfly theory with double line

if we extend $PB$ and $PC$ to intersect $AC$ and $AB$ in $X$ and $Y$ then $XY$ is parallel to $EF$

can anyone show me how to prove this using barycentric coordinates

Let $BP \cap AC=K$, $CP \cap AB=L$, Let $\ell$ be the parallel to $DM$ through $K$ and Let $\ell \cap BC=X$, and let $KL \cap BC=Y$ $$-1=(F,E;D, \infty_{DM}) \overset{K}{=} (C,B;D,X) \text{ but, } -1=(C,B;D,Y) \Longrightarrow \boxed{X \equiv Y} \implies LK||DM$$Then, $$-1=(C,B;D,X) \overset{L}{=} (N,M;D, \infty_{DM} ) \implies \boxed{DM=DN}$$