Proof that fast broadcasts have to target larger cycles first
Let $b_0$ be the minimal broadcasting time of an information from vertex $v$ to the cycle $C_0$ and $b_1$ be the minimal broadcasting time of an information from vertex $v$ to the cycle $C_1$.

Remark that since $C_0$ and $C_1$ are cycles the fastest way to broad cast is for $v$ to broadcast first one of his neighbor and then the other.

Since $C_0$ and $C_1$ are only connected by $v$ and $v$ already have the information we know that spreading the information through $C_i$ will not improve the minimal broadcasting time for $C_{1-i}$.

3 cases (first one developed the others just ideas):


$b_0&gt;=b_1+2$ (hence $|C_1|&gt;|C_2|+3$) if $v$ start broadcasting his two neighbor in $C_0$ and then those of $C_1$. The time the information reach the last vertex in $C_0$ will be $b_0$ and the last vertex in $C_1$ $b_1+2$ hence $b_0$ in total. And you can't do better by definition of $b_0$
$b_0=b_1+1$ if you start twice by $1$ you will do at best $b_0+2$ and you do better by choosing $0$
$b_0=b_1$ it doesn't really matter but if the parity of the two cycle is different you may want to start by the odd one (or even not sure :D)