Prove that four points are concyclic

Question

We are given four circles : $C_{1}, C_{2},C_{3},C_{4},$

Such as :

$\;C_{1}\;$ is concurrent to $\;C_{4}\;$ in $A\;$ & $\;A'$

$\;C_{1}\;$ is concurrent to $\;C_{2}\;$ in $B\;$ & $\;B'$

$\;C_{2}\;$ is concurrent to $\;C_{3}\;$ in $C\;$ & $\;C'$

$\;C_{3}\;$ is concurrent to $\;C_{4}\;$ in $D\;$ & $\;D'$

I have to show that $\;A,B,C,D\;$ lie on a common circle if and only if $\;A',B',C',D'\;$ are concyclic

So it can be more clear

Answer 1

Hint: Direct angle chasing.
Show that $ \angle BCD + \angle BAD = \angle C'D'A' + \angle C'B'A'$

Hence, $ABCD$ is concyclic iff $$ \angle BCD + \angle BAD = 180^\circ$ iff $\angle C'D'A' + \angle C'B'A' = 180^\circ$ iff $A'B'C'D'$ is concyclic.

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For the details

$\angle BCD = 360^\circ - \angle C'CD - \angle C'CB = \angle C'D'D + \angle C'B'B$
Likesiwse $\angle BAD = \angle A'D'D + \angle A'B'B$.
Sum up these 2 angles to get $\angle C'D'A' + \angle C'B'A'$