Given triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully.

 

Through the investigation we can assure two resuls the above for acute triangle.

Let me show an example.

 

PROOF)

We can represent an area of the triangle above as three defferent formulas.

Area of ABC = (BC * AD)/2

Area of ABC = (CA * BE)/2

Area of ABC = (AB * CF)/2

And

Area ABC = Area HBC + Area HCA + Area HAB

Area HBC = (BC*HD)/2

Area HCA = (CA*HE)/2

Area HAB = (AB*HF)/2

 

Therefore

2 * AreaABC = BC*HD + CA*HE + AB*HF

BC*AD = BC*HD + CA*HE + AB*HF

Now we can get the first result by the following steps

1 = (BC*HD)/(BC*AD) + (CA*HE)/(BC*AD) + (AB*HF)/(BC*AD)

By three different area formulas BC*AD = CA*BE = AB*CF

1 = (BC*HD)/(BC*AD) + (CA*HE)/(CA*BE) + (AB*HF)/(AB*CF)

That is,

HD/AD + HE/BE + HF/CF = 1

 

Second proof can be directly induced from the first result.

HD = AD - AH

HE = BE - BH

HF = CF - CH

Let's put these equations into the first result.

(AD- AH)/AD + (BE - BH)/BE + (CF-CH)/CF = 1

1 - AH/AD + 1 - BH/BE + 1 - CH/CF = 1

3 - (AH/AD + BH/BE + CH/CF) = 1

That is,

AH/AD + BH/BE + CH/CF = 2

Further study

What if the triangle is right?

Orthocenter is to be a point B and so our B, D, F, H are coincidence.

Since HD=0, HF=0, BE=HE,

HD/AD + HE/BE + HF/CF = 1

And since BH=0, AD=AH, CF=CH,

AH/AD + BH/BE + CH/CF = 2

 

What if the triangle is obtuse?

More detail figure is the following

From this figure we can know that our two results are not satisfied for the obtuse.

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