If $A + B = R$ and $A+C=R$ then $A+BC=R$ and $A \cap BC = ABC$

Hello I'm trying to prove this problem: Let $R$ be a commutative ring and $A,B,C \subseteq R$ ideals such that $A+B=R=A+C$ then $A+BC=R$ and $A \cap BC = ABC$.

The last statement is easy to prove once the first has been proved, since $1 \in R = A+BC$ hence $1=a+y$ for some $a \in A, y \in BC$ so for each $x \in A \cap (BC)$ we have that $x=x*1 = x*(a+y)= xa + xy=ax+xy$ because $R$ is commutative. Since $ax \in A(BC)$ and $xy \in A (BC)$ then $x \in A(BC) = ABC$.

And as $ABC = A(BC) \subseteq A \cap (BC)$ we have finished.

I need some help to prove the first statement, I don't see clearly how to prove that. I would appreacite any help.

We know that $R=A+B$, and so there are $a_1\in A, b\in B$ such that $1=a_1+b$. Also, $R=A+C$ and so $1=a_2+c$ for some $a_2\in A, c\in C$. And then:

$1=1\cdot 1=(a_1+b)(a_2+c)=a_1a_2+a_2b+a_1c+bc$

Since $A$ is an ideal we have $a_1a_2+a_2b+a_1c\in A$, and so:

$1=(a_1a_2+a_2b+a_1c)+bc\in A+BC$

If $a+b=1$ and $a'+c=1$, then $$ 1=a'+(a+b)c=a'+ac+bc.$$