Geometric Proof: Proving Independence of DE + DF in Isosceles Triangle ABC

[DotKite]

Homework Statement

In triangle ABC, AB = AC, and D,E,F are points on the interiors of sides BC,AB,AC respectively, such that DE perpendicular to AB and DF perpendicular to AC. Prove that the value of DE + DF is independent of the location of D

Homework Equations

So far we have all the tools of neutral geometry and non neutral parallelism. We have not covered similarity yet

The Attempt at a Solution

Ok so I guess a good approach would be to consider triangle ABC with D in one location and then another and show there is no change in DE + DF. However I do not have a clue in how to proceed with this. Tried using the fact that triangle ABC is isosceles therefore the base angles are equal, but don't really know where to go with that either

 

[tiny-tim]

In triangle ABC, AB = AC, and D,E,F are points on the interiors of sides BC,AB,AC respectively, such that DE perpendicular to AB and DF perpendicular to AC. Prove that the value of DE + DF is independent of the location of D

I don't think that's true.

Suppose D is either B or the midpoint of BC.

Then the perpendicular from B to AC would have to be twice the perpendicular from that midpoint.

But it isn't, is it?

 

[LCKurtz]

I don't think that's true.

Suppose D is either B or the midpoint of BC.

Then the perpendicular from B to AC would have to be twice the perpendicular from that midpoint.

But it isn't, is it?

Yes, I think it is. Draw the midline parallel to AC. It will bisect BF.

[Edit, added]: It's true alright, and easy enough to prove analytically. The sum of DE and DF comes out equal to the length of the base divided by the length of one of the equal legs, independent of the location of D. Not that any of this helps the OP.

 

[dirk_mec1]

@OP, can you post a picture it will truly help.

 

[tiny-tim]

Yes, I think it is. Draw the midline parallel to AC. It will bisect BF.

[Edit, added]: It's true alright, and easy enough to prove analytically. The sum of DE and DF comes out equal to the length of the base divided by the length of one of the equal legs, independent of the location of D. Not that any of this helps the OP.

not following you

(how can the sum of two lengths equal the ratio of two lengths, ie a number?)

anyway, don't forget that you can't use most of the usual theorems about parallel lines

 

[LCKurtz]

not following you

(how can the sum of two lengths equal the ratio of two lengths, ie a number?)

anyway, don't forget that you can't use most of the usual theorems about parallel lines

Sorry, I left out a factor. The sum of the two lengths is equal to $\frac {2ha}{\sqrt{a^2+h^2}}$ where $h$ is the height of the triangle and $2a$ is the length of the base. So the units are area/length = length. But, as I said, not the proof the OP is looking for.