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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