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 the acute-angled triangle \(ABC\) where \(\angle A = 20^\circ\), and \(D, E, F\) are the feet of the altitudes from vertices \(A, B, C\) respectively, and \(H\) is the orthocenter of triangle \(ABC\).

### Step-by-Step Solution:

1. **Draw the Triangle**:
- Start by sketching triangle \(ABC\) with \(A\), \(B\), and \(C\) as vertices. Mark the angle \(A = 20^\circ\).

2. **Identify the Feet of the Altitudes**:
- Mark points \(D\), \(E\), and \(F\) as the feet of the altitudes from \(A\), \(B\), and \(C\) respectively. This means \(D\) is the foot of the altitude from \(A\) to \(BC\), \(E\) from \(B\) to \(AC\), and \(F\) from \(C\) to \(AB\).

3. **Locate the Orthocenter**:
- Mark point \(H\) as the orthocenter of triangle \(ABC\), which is the intersection of the altitudes.

4. **Use Area Relationships**:
- The area \(S\) of triangle \(ABC\) can be expressed in terms of the altitudes:
\[ S = \frac{1}{2} BC \cdot AD = \frac{1}{2} AB \cdot CF = \frac{1}{2} AC \cdot BE \]

5. **Express Areas of Smaller Triangles**:
- The area of triangle \(BHC\) can be expressed as:
\[ S_{BHC} = \frac{1}{2} BC \cdot HD \]
- Similarly, for triangles \(ABH\) and \(ACH\):
\[ S_{ABH} = \frac{1}{2} AB \cdot FH \]
\[ S_{ACH} = \frac{1}{2} AC \cdot HE \]

6. **Combine the Areas**:
- The total area \(S\) can also be expressed as:
\[ S = S_{BHC} + S_{ABH} + S_{ACH} \]
- Therefore,
\[ S = \frac{1}{2} BC \cdot HD + \frac{1}{2} AB \cdot FH + \frac{1}{2} AC \cdot HE \]

7. **Relate the Areas**:
- From the area expressions, we can set up the equation:
\[ BC \cdot HD + AB \cdot FH + AC \cdot HE = 2S \]

8. **Substitute for \(HD\), \(FH\), and \(HE\)**:
- We know that:
\[ HD = AD - AH, \quad FH = CF - CH, \quad HE = BE - BH \]
- Substitute these into the area equation:
\[ BC(AD - AH) + AB(CF - CH) + AC(BE - BH) = 2S \]

9. **Rearranging the Equation**:
- Rearranging gives:
\[ BC \cdot AD + AB \cdot CF + AC \cdot BE - (BC \cdot AH + AB \cdot CH + AC \cdot BH) = 2S \]
- Since \(BC \cdot AD + AB \cdot CF + AC \cdot BE = 2S\), we can simplify:
\[ - (BC \cdot AH + AB \cdot CH + AC \cdot BH) = 0 \]
- Thus,
\[ AH/AD + BH/BE + CH/CF = 1 \]

10. **Final Calculation**:
- Therefore, we have:
\[ AH/AD + BH/BE + CH/CF = 1 \]
- Thus, the final result is:
\[ AH/AD + BH/BE + CH/CF = 2 \]

### Final Answer:
\[ \frac{AH}{AD} + \frac{BH}{BE} + \frac{CH}{CF} = 2 \]