In an acute angled triangle ABC, `angleA=20^(@)`, let DEF be the feet of altitudes through A, B, C respectively and H is the orthocentre of `DeltaABC`. Find `(AH)/(AD)+(BH)/(BE)+(CH)/(CF)`.

Answer

To solve the problem, we need to find the value of \((AH)/(AD) + (BH)/(BE) + (CH)/(CF)\) in an acute-angled triangle \(ABC\) where \(\angle A = 20^\circ\) and \(D, E, F\) are the feet of the altitudes from \(A, B, C\) respectively, and \(H\) is the orthocenter of triangle \(ABC\).

### Step-by-Step Solution:

1. **Understanding the Area of Triangle**:
The area \(S\) of triangle \(ABC\) can be expressed in three different ways using the altitudes:
\[ S = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times AC \times BE = \frac{1}{2} \times AB \times CF \]

2. **Expressing Area in Terms of Orthocenter**:
The area \(S\) can also be expressed as the sum of the areas of triangles \(BHC\), \(ABH\), and \(ACH\):
\[ S = \text{Area}(BHC) + \text{Area}(ABH) + \text{Area}(ACH) \]

3. **Using the Area Formulas**:
Using the base-height formula for each of these triangles:
- Area of triangle \(BHC = \frac{1}{2} \times BC \times HD\)
- Area of triangle \(ABH = \frac{1}{2} \times AB \times HF\)
- Area of triangle \(ACH = \frac{1}{2} \times AC \times HE\)

4. **Setting Up the Equations**:
We can express the area \(S\) in terms of \(AD\), \(BE\), and \(CF\):
\[ S = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times AB \times CF = \frac{1}{2} \times AC \times BE \]

5. **Relating Heights to Altitudes**:
From the area expressions, we can relate the heights to the altitudes:
\[ 2S = BC \times AD \Rightarrow AD = \frac{2S}{BC} \]
Similarly, we can find:
\[ BE = \frac{2S}{AC}, \quad CF = \frac{2S}{AB} \]

6. **Substituting into the Original Expression**:
Now substituting these into the expression we want to evaluate:
\[ \frac{AH}{AD} + \frac{BH}{BE} + \frac{CH}{CF} \]

7. **Using the Relationships of Heights**:
We know that:
- \(HD = AD - AH\)
- \(HF = CF - CH\)
- \(HE = BE - BH\)

Therefore, we can express:
\[ AH = AD - HD, \quad BH = BE - HF, \quad CH = CF - HE \]

8. **Final Calculation**:
After substituting these values into the expression, we find:
\[ \frac{AH}{AD} + \frac{BH}{BE} + \frac{CH}{CF} = 3 - 1 = 2 \]

### Conclusion:
Thus, the final result is:
\[ \frac{AH}{AD} + \frac{BH}{BE} + \frac{CH}{CF} = 2 \]