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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.

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.

]]>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)

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

]]>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??

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