From your Venn diagram you can see that the $bc$ term on the lefthand side adds to ab+a'b' only what’s in $b$ and $c$ and not in $a$: any part of $bc$ that’s in $a$ is already taken care of by the $ab$ term. Similarly, the a'c term on the righthand side adds only what’s in $c$, not in $a$, and in $b$: any part of a'c that’s not in $b$ is already taken care of by the a'b' term. This suggests trying to prove that both sides are equal to ab+a'b'+a'bc.
For the lefthand side, remember that we noticed pictorially that $bc$ could be replaced by a'bc because the $abc$ part of it was already covered by $ab$. This suggests that we should try splitting it into those two parts:
\begin{align*} ab+a'b'+bc&=ab+a'b'+(a+a')bc\\ &=ab+a'b'+abc+a'bc\\ &=(ab+abc)+a'b'+a'bc\\ &=ab+a'b'+a'bc\;. \end{align*}
For the righthand side we saw that a'c could be replaced by a'bc because the a'b'c part was already covered by a'b', so this time I split a'c into a'bc and a'b'c:
\begin{align*} ab+a'b'+a'c&=ab+a'b'+a'(b+b')c\\ &=ab+a'b'+a'bc+a'b'c\\ &=ab+(a'b'+a'b'c)+a'bc\\ &=ab+a'b'+a'bc\;. \end{align*}
The result now follows immediately.