Consider a right-angled triangle with integer-valued sides $a<b<c$ where $a,b,c$ are pairwise co-prime. Let $d=c-b$ . Suppose $d$ divides $a$ . Then 

[b](a)[/b]  Prove that $d\leqslant 2$. 

[b](b)[/b]  Find all such triangles (i.e. all possible triplets $a,b,c$) with perimeter less than $100$ .
1. **Prove that \( d \leq 2 \)**

   Given that \( a, b, c \) are the sides of a right-angled triangle with \( a^2 + b^2 = c^2 \) and \( d = c - b \). We are also given that \( d \) divides \( a \).

   Since \( d = c - b \), we can write:
   \[
   c = b + d
   \]
   Substituting \( c = b + d \) into the Pythagorean theorem:
   \[
   a^2 + b^2 = (b + d)^2
   \]
   Expanding and simplifying:
   \[
   a^2 + b^2 = b^2 + 2bd + d^2
   \]
   \[
   a^2 = 2bd + d^2
   \]
   Since \( d \) divides \( a \), let \( a = kd \) for some integer \( k \). Substituting \( a = kd \) into the equation:
   \[
   (kd)^2 = 2bd + d^2
   \]
   \[
   k^2d^2 = 2bd + d^2
   \]
   Dividing through by \( d^2 \):
   \[
   k^2 = \frac{2b}{d} + 1
   \]
   Rearranging:
   \[
   \frac{2b}{d} = k^2 - 1
   \]
   Since \( \frac{2b}{d} \) must be an integer, \( k^2 - 1 \) must be divisible by \( d \). Therefore, \( d \) must be a divisor of \( k^2 - 1 \).

   Now, since \( a, b, c \) are pairwise co-prime, \( d \) must be small. We will check the possible values of \( d \):

   - If \( d = 1 \):
     \[
     \frac{2b}{1} = k^2 - 1 \implies 2b = k^2 - 1
     \]
     Since \( b \) is an integer, \( k^2 - 1 \) must be even, which is always true for any integer \( k \).

   - If \( d = 2 \):
     \[
     \frac{2b}{2} = k^2 - 1 \implies b = \frac{k^2 - 1}{2}
     \]
     For \( b \) to be an integer, \( k^2 - 1 \) must be even, which is true for any integer \( k \).

   - If \( d \geq 3 \):
     \( d \) must divide \( k^2 - 1 \). However, since \( a, b, c \) are pairwise co-prime, larger values of \( d \) would imply that \( a \) and \( b \) share a common factor, which contradicts the given condition.

   Therefore, \( d \leq 2 \).

   \(\blacksquare\)

2. **Find all such triangles with perimeter less than \( 100 \)**

   We will consider the cases \( d = 1 \) and \( d = 2 \) separately.

   **Case 1: \( d = 1 \)**

   Here, \( c = b + 1 \). Substituting into the Pythagorean theorem:
   \[
   a^2 + b^2 = (b + 1)^2
   \]
   \[
   a^2 + b^2 = b^2 + 2b + 1
   \]
   \[
   a^2 = 2b + 1
   \]
   Since \( a \) is an integer, \( 2b + 1 \) must be a perfect square. Let \( 2b + 1 = k^2 \) for some integer \( k \):
   \[
   b = \frac{k^2 - 1}{2}
   \]
   Since \( b \) must be an integer, \( k \) must be odd. Let \( k = 2m + 1 \) for some integer \( m \):
   \[
   b = \frac{(2m + 1)^2 - 1}{2} = 2m^2 + 2m
   \]
   \[
   a = 2m + 1
   \]
   \[
   c = b + 1 = 2m^2 + 2m + 1
   \]
   The perimeter is:
   \[
   a + b + c = (2m + 1) + (2m^2 + 2m) + (2m^2 + 2m + 1) = 4m^2 + 6m + 2
   \]
   We need \( 4m^2 + 6m + 2 < 100 \):
   \[
   4m^2 + 6m + 2 < 100
   \]
   \[
   4m^2 + 6m - 98 < 0
   \]
   Solving the quadratic inequality:
   \[
   m = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 4 \cdot (-98)}}{2 \cdot 4} = \frac{-6 \pm \sqrt{36 + 1568}}{8} = \frac{-6 \pm \sqrt{1604}}{8}
   \]
   \[
   \sqrt{1604} \approx 40.05
   \]
   \[
   m = \frac{-6 \pm 40.05}{8}
   \]
   \[
   m \approx 4.25 \quad \text{or} \quad m \approx -5.75
   \]
   Since \( m \) must be a non-negative integer, \( m \) can be \( 0, 1, 2, 3, 4 \).

   - For \( m = 0 \):
     \[
     a = 1, \quad b = 0, \quad c = 1 \quad \text{(not valid as \( b \) must be positive)}
     \]

   - For \( m = 1 \):
     \[
     a = 3, \quad b = 4, \quad c = 5
     \]
     Perimeter: \( 3 + 4 + 5 = 12 \)

   - For \( m = 2 \):
     \[
     a = 5, \quad b = 12, \quad c = 13
     \]
     Perimeter: \( 5 + 12 + 13 = 30 \)

   - For \( m = 3 \):
     \[
     a = 7, \quad b = 24, \quad c = 25
     \]
     Perimeter: \( 7 + 24 + 25 = 56 \)

   - For \( m = 4 \):
     \[
     a = 9, \quad b = 40, \quad c = 41
     \]
     Perimeter: \( 9 + 40 + 41 = 90 \)

   **Case 2: \( d = 2 \)**

   Here, \( c = b + 2 \). Substituting into the Pythagorean theorem:
   \[
   a^2 + b^2 = (b + 2)^2
   \]
   \[
   a^2 + b^2 = b^2 + 4b + 4
   \]
   \[
   a^2 = 4b + 4
   \]
   Since \( a \) is an integer, \( 4b + 4 \) must be a perfect square. Let \( 4b + 4 = k^2 \) for some integer \( k \):
   \[
   b = \frac{k^2 - 4}{4}
   \]
   Since \( b \) must be an integer, \( k^2 - 4 \) must be divisible by 4. Let \( k = 2m \) for some integer \( m \):
   \[
   b = \frac{(2m)^2 - 4}{4} = m^2 - 1
   \]
   \[
   a = 2m
   \]
   \[
   c = b + 2 = m^2 + 1
   \]
   The perimeter is:
   \[
   a + b + c = 2m + (m^2 - 1) + (m^2 + 1) = 2m^2 + 2m
   \]
   We need \( 2m^2 + 2m < 100 \):
   \[
   m^2 + m < 50
   \]
   Solving the quadratic inequality:
   \[
   m^2 + m - 50 < 0
   \]
   \[
   m = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-50)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 200}}{2} = \frac{-1 \pm \sqrt{201}}{2}
   \]
   \[
   \sqrt{201} \approx 14.18
   \]
   \[
   m = \frac{-1 \pm 14.18}{2}
   \]
   \[
   m \approx 6.59 \quad \text{or} \quad m \approx -7.59
   \]
   Since \( m \) must be a non-negative integer, \( m \) can be \( 0, 1, 2, 3, 4, 5, 6 \).

   - For \( m = 0 \):
     \[
     a = 0, \quad b = -1, \quad c = 1 \quad \text{(not valid as \( b \) must be positive)}
     \]

   - For \( m = 1 \):
     \[
     a = 2, \quad b = 0, \quad c = 2 \quad \text{(not valid as \( b \) must be positive)}
     \]

   - For \( m = 2 \):
     \[
     a = 4, \quad b = 3, \quad c = 5
     \]
     Perimeter: \( 4 + 3 + 5 = 12 \)

   - For \( m = 3 \):
     \[
     a = 6, \quad b = 8, \quad c = 10 \quad \text{(not pairwise co-prime)}
     \]

   - For \( m = 4 \):
     \[
     a = 8, \quad b = 15, \quad c = 17
     \]
     Perimeter: \( 8 + 15 + 17 = 40 \)

   - For \( m = 5 \):
     \[
     a = 10, \quad b = 24, \quad c = 26 \quad \text{(not pairwise co-prime)}
     \]

   - For \( m = 6 \):
     \[
     a = 12, \quad b = 35, \quad c = 37
     \]
     Perimeter: \( 12 + 35 + 37 = 84 \)

   Therefore, the valid triangles with perimeter less than 100 are:
   \[
   (3, 4, 5), (5, 12, 13), (7, 24, 25), (9, 40, 41), (8, 15, 17), (12, 35, 37)
   \]

The final answer is \( \boxed{ (3, 4, 5), (5, 12, 13), (7, 24, 25), (9, 40, 41), (8, 15, 17), (12, 35, 37) } \)