The opposite sides of the reentrant hexagon $AFBDCE$ intersect at the points $K,L,M$ (as shown in the figure). It is given that $AL = AM = a, BM = BK = b$, $CK = CL = c, LD = DM = d, ME = EK = e, FK = FL = f$. 
[img]http://imgur.com/LUFUh.png[/img]
$(a)$ Given length $a$ and the three angles $\alpha, \beta$ and $\gamma$ at the vertices $A, B,$ and $C,$ respectively, satisfying the condition $\alpha+\beta+\gamma<180^{\circ}$, show that all the angles and sides of the hexagon are thereby uniquely determined.
$(b)$ Prove that
\[\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{d}\]
Easier version of $(b)$. Prove that
\[(a + f)(b + d)(c + e)= (a + e)(b + f)(c + d)\]
### Part (a)
Given the lengths \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\) and the angles \(\alpha\), \(\beta\), and \(\gamma\) at vertices \(A\), \(B\), and \(C\) respectively, we need to show that all the angles and sides of the hexagon are uniquely determined.

1. **Determine the angles at vertices \(D\), \(E\), and \(F\):**
   Since the sum of the internal angles of a hexagon is \(720^\circ\), and given \(\alpha + \beta + \gamma < 180^\circ\), we can find the remaining angles:
   \[
   \delta = 180^\circ - \alpha, \quad \epsilon = 180^\circ - \beta, \quad \zeta = 180^\circ - \gamma
   \]
   These angles are uniquely determined by \(\alpha\), \(\beta\), and \(\gamma\).

2. **Determine the sides of the hexagon:**
   The sides \(AL\), \(AM\), \(BM\), \(BK\), \(CK\), \(CL\), \(LD\), \(DM\), \(ME\), \(EK\), \(FK\), and \(FL\) are given as \(a\), \(a\), \(b\), \(b\), \(c\), \(c\), \(d\), \(d\), \(e\), \(e\), \(f\), and \(f\) respectively. These lengths are uniquely determined.

3. **Determine the intersection points \(K\), \(L\), and \(M\):**
   The points \(K\), \(L\), and \(M\) are determined by the intersection of the opposite sides of the hexagon. Given the lengths and angles, these points are uniquely determined by the geometry of the hexagon.

Thus, all the angles and sides of the hexagon are uniquely determined by the given lengths and angles.

### Part (b)
We need to prove the following identity:
\[
\frac{1}{a} + \frac{1}{c} = \frac{1}{b} + \frac{1}{d}
\]

1. **Using the circumradius \(R\) of \(\triangle KLM\):**
   \[
   a = \frac{ML}{2} \csc\left(\frac{\alpha}{2}\right) = R \sin(\angle MKL) \csc\left(\frac{\alpha}{2}\right)
   \]
   \[
   d = \frac{ML}{2} \csc\left(\frac{\angle D}{2}\right) = R \sin(\angle MKL) \csc\left(\frac{\angle D}{2}\right)
   \]

2. **Expressing the difference:**
   \[
   \frac{1}{d} - \frac{1}{a} = \frac{1}{R} \left( \sin\left(\frac{\angle D}{2}\right) - \sin\left(\frac{\alpha}{2}\right) \right) \csc(\angle MKL)
   \]
   \[
   = \frac{2}{R} \cos\left(\frac{\angle D + \alpha}{4}\right) \sin\left(\frac{\angle D - \alpha}{4}\right) \csc(\angle MKL)
   \]

3. **Simplifying using trigonometric identities:**
   \[
   \cos\left(\frac{\angle D + \alpha}{4}\right) = \sin(\angle MKL)
   \]
   \[
   \frac{1}{d} - \frac{1}{a} = \frac{2}{R} \sin\left(\frac{\angle D - \alpha}{4}\right) = \frac{2}{R} \sin\left(\frac{\pi - \alpha - \beta - \gamma}{6}\right)
   \]

4. **Similarly for \(e\) and \(b\):**
   \[
   \frac{1}{e} - \frac{1}{b} = \frac{2}{R} \sin\left(\frac{\pi - \alpha - \beta - \gamma}{6}\right) = \frac{1}{d} - \frac{1}{a}
   \]

5. **Combining the results:**
   \[
   \frac{1}{a} + \frac{1}{e} = \frac{1}{b} + \frac{1}{d}
   \]

### Easier Version
We need to prove:
\[
(a + f)(b + d)(c + e) = (a + e)(b + f)(c + d)
\]

1. **Using the given identity:**
   \[
   \frac{a + e}{ae} = \frac{b + d}{bd}
   \]
   \[
   \frac{b + f}{bf} = \frac{c + e}{ce}
   \]
   \[
   \frac{c + d}{cd} = \frac{a + f}{af}
   \]

2. **Combining the fractions:**
   \[
   \frac{(a + f)(b + d)(c + e)}{(a + e)(b + f)(c + d)} = \frac{afbdce}{aebfcd} = 1
   \]

Thus, the identity holds.

The final answer is \( \boxed{ \frac{1}{a} + \frac{1}{c} = \frac{1}{b} + \frac{1}{d} } \)