Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**triangle-maniac****Guest**

In triangle ABC which has acute angles Heights are cut across in point H.

A Straight line goes through point H and it crosses AC in point D and BC in point E. A second straight line goes through H and makes a right angle with DE and crosses AB in point F.

Prove that: DH/HE = AF/FB

Can someone help??

**ryos****Member**- Registered: 2005-08-04
- Posts: 394

That problem statement is greek to me. Is that how they gave it to you? I'm sorry.

El que pega primero pega dos veces.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

A sketch would really help ... was there any diagram with it?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**juukii****Guest**

http://juukii.w.interia.pl/iiii.bmp

I've made a diagram that suits but I can't figure it out

**triangle-maniac****Guest**

Yes this is exactlly the same diagram which they gave to me

**triangle-maniac****Guest**

Can someone help now please??

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

The diagram took a long time to load for me, so I have copied it here (click to enlarge)

OK, it LOOKS like DH/HE and AF/FB might be linked, but I can't figure out how to prove it yet.

One clue is that if the line DE runs ALONG one of the "height" lines, then you can see that the ratio of DH/HE is the same as AF/FB (second diagram)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**triangle-maniac****Guest**

Please If u could help me prove it I would be veary,veary happy it's veary important for me...

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

I have sat and looked at it for a while, but all I have come up with is what I mentioned.

If HE is parallel to HB (in fact runs along it), then HF is parallel to AD, and the ratios are exactly the same. (The two triangles ADB and FHB are *similar* triangles.)

This would also be true if the DE line were aligned along the AH line.

I am hoping another member can take this further, because I have to leave right now, but I will come back later to see how this is going.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

Still haven't solved it, but have some more ideas.

First Image: If DE is parallel to AB, then the ratio is easy to prove (AFB can be seen to be an enlargement of DHE)

Second Image: If we now rotate DE a little, can we then prove that the ratio stays the same? I hope so.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

This is just a thought, but in the second picture, could we move point C to the right a little bit and make it look like the first picture again? If that can happen without changing the ratio of AD:DC, then it could be done no matter where D is and so it would sort of be a proof, albeit a very sloppy one.

Why did the vector cross the road?

It wanted to be normal.

Offline

**triangle-maniac****Guest**

I can't belive it's such a hard problem..IS there someone who can help??

Please

Pages: **1**