triangle problem

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jred

290 posts

jred

#1 h Jun 29, 2013, 10:49 AM • 2 Y

Y by Adventure10, Mango247

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.

This post has been edited 1 time. Last edited by jred, Jun 29, 2013, 10:59 AM

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War-Hammer

670 posts

War-Hammer

#2 h Jun 29, 2013, 10:57 AM • 2 Y

Y by Adventure10, Mango247

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 $DM=DN$ ???

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jred

290 posts

jred

#3 h Jun 29, 2013, 11:03 AM • 2 Y

Y by Adventure10, Mango247

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.

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Hantaehee

99 posts

Hantaehee

#4 h Jun 29, 2013, 2:30 PM • 2 Y

Y by Adventure10, Mango247

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

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GMOH

52 posts

GMOH

#5 h Feb 20, 2018, 8:40 AM • 2 Y

Y by Adventure10, Mango247

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

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AlastorMoody

2125 posts

AlastorMoody

#6 h Nov 11, 2018, 3:11 AM • 2 Y

Y by Adventure10, Mango247

can anyone show me how to prove this using barycentric coordinates

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AlastorMoody

2125 posts

AlastorMoody

#7 h Feb 23, 2019, 1:36 PM • 2 Y

Y by Adventure10, Mango247

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, $-1=(C,B;D,X) \overset{L}{=} (N,M;D, \infty_{DM} ) \implies \boxed{DM=DN}$

This post has been edited 2 times. Last edited by AlastorMoody, Feb 23, 2019, 1:37 PM

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