Geometry Challenge: Prove Perpendicularity of HE to AB

[krypt]

Hello, I'm new here (to posting not to the forums).
The following is a challange proposed by our math teacher.
The work is due tomorrow morning.

1. The problem statement/known data:

ABC is a right triangle. (BAC=90°)
[AH] is an "altitude" (AH_|_BC)
I is the center of [BC]
J is the center of [AH]
Starting from A, lead a perpendicular to (IJ) which intersects (BC). The intersection is F.
(JF) intersects (AB), the intersection is E.
Prove that (HE) is perpendicular to (AB).

2. The attempt at a solution

I have tried working in the FHXA rectangle (X being H's orthogonal projection on (AC), my intuition keeps taking me there but I'm still having no luck solving it.



Thank you for your help,
I do appreaciate it.

I am sorry for the ill terms I might have used as I am French educated.

 

[dynamicsolo]

I think you're going to need to clarify certain requirements or draw a picture of what you've done. I'm interpreting this to mean that A is the right angle in the right triangle, I is the midpoint of the hypotenuse BC, and J is the midpoint of the constructed altitude AH.

What does this statement mean? "Starting from A, lead a perpendicular to (IJ) which intersects (BC)." A line starting from point A isn't going to cross IJ at right angles.

"The intersection is F. (JF) intersects (AB), the intersection is E." And F would have to be between H and C in order for the line through J and F to intersect AB at all.

It is rather unclear where the rectangle FHXA is. Could you describe the construction of F and E more clearly, or include a diagram, please?

 

[krypt]

Thank you for your reply.
Sorry for the confusion.

Hope this helps


please find the pic on h**p://ftp.screensint.com/math.jpg

(replace h**P with http)

 

[dynamicsolo]

Your approach might work, though it seems like it will be somewhat roundabout. Basically, you'd like to prove in some way that triangle BEH is similar to triangle BAC, which would immediately lead to the conclusion that angle BEH is congruent to angle BAC, and therefore is a right angle.

Your triangle AXH is similar to triangle BAC; it would be nice to show that, in turn, triangle HEA is congruent to AXH, because that would also prove that AEH, and thus BEH, is a right angle. But looks may be deceiving... There probably is a way to get down that road, but I think another route may be more fruitful.

If you start marking angles in the constructed triangles, it is pretty straightforward to show that angles FAJ and FIJ are congruent (look at vertical angles and the angles in the various known right triangles), and thus that triangles FAJ and FIJ are congruent. I'll also mention that right triangle HIJ seems to be a sort of Rosetta Stone in establishing a numerical relation between certain angular measures that is then helpful elsewhere.

I haven't pinned everything down yet and cleaned up the demonstration to make it look elegant (rather than the stumbling about in the diagram that it currently is), but it appears that you should be able to prove one of two things:

either

angles ABC and EBH are congruent and angles BHE and BCA are congruent, and thus triangles BEH abd BAC are similar, hence angle BEH is a right angle

or

angles EHA and HAC are congruent, and thus constitute a pair of alternate interior angles, hence EH is parallel to AC, making AEH and BEH right angles.