geometry proof

[Guest]

Hi there,

I am trying to prove the following:

Triangle ABC is acute, the bases of the altitudes at B and C are E and F respectively.
Let A' be the midpoint of BC. Prove that Triangle A'EF is isoceles.


I drew an example of this on sketchpad, here:

By inspection you can see that |A'F| = |A'E| so angle EFA' = angle FEA'

I am not sure where to start, I was thinking of drawing the parallel to AC, AB, BC that pass through points B, C, and A respectively, creating an outer triangle that would be similar to triangle ABC and I think it would create the perpendicular bisectors of this new bigger triangle but I couldnt see any relation between this and triangle A'EF, only ABC

Any thoughts? [/img]

 

[stapel]

You have a lot of measurements on your drawing. Are they givens, or are they just the specifics of this particular illustration?

Eliz.

 

[Guest]

You have a lot of measurements on your drawing. Are they givens, or are they just the specifics of this particular illustration?

Eliz.

No it was just a single instance of what it could look like. I actually figured it out by drawing a circle with diameter |BC| centered at A'

since the altitudes form 90 degree angles which also subtend the same arc as the diameter, A'E and A'F are radii

therefore the triangle is isoceles.