A problem by OJAS JAIN

Given the first equation:

$a = b - c ...(1)$

and the second one:

$b = c - d ... (2)$

If we use substitution for the first and second equation, we'll get:

$a = b - c$

$a = c - d - c$

$\boxed{a = -d} ... (4)$

Given the third equation:

$c = d - a ... (3)$

If we use substitution for the second and third equation, we'll get:

$b = c - d$

$b = d - a - d$

$\boxed{b = -a = d} ... (5)$

If we use substitution for the third and fourth equation, we'll get:

$c = d - a$

$c = d + d$

$\boxed{c = 2d} ... (6)$

If we put all equation into the main equation, we'll get:

$\frac {a}{b} + \frac {b}{c} + \frac {c}{d} + \frac {d}{a}$

$= \frac {-d}{d} + \frac {d}{2d} + \frac {2d}{d} + \frac {d}{-d}$

$= -1 + \frac {1}{2} + 2 - 1$

$= \boxed{\frac {1}{2}}$

$\boxed{TADAAAA}$