$\newcommand{\r}[1]{\mathrel{#1}}$ First, a few definitions. Given a lattice $L$, a congruence on $L$ is an equivalence relation $\theta$, compatible with the lattice operations, i.e. if $x_1\r{\theta}x_2$ and $y_1\r{\theta}y_2$, then $x_1\wedge x_2\r{\theta}y_1\wedge y_2$ and $x_1\vee x_2\r{\theta}y_1\vee y_2$. The congruences on $L$ form a lattice, with meet being intersection and join being the equivalence envelope.
I'm trying to prove that this lattice of congruences is distributive. So let's take three congruences $\theta_1,\theta_2,\theta_3$ and attempt to prove that $\theta_1\cap(\theta_2\vee\theta_3)\subseteq (\theta_1\cap\theta_2)\vee(\theta_1\cap\theta_3)$
For a start, let's try to prove something easier which should lead me to the general idea. Take elements $x,y,z$ such that $x\r{\theta_1}y,x\r{\theta_2}z$ and $z\r{\theta_3}y$. We want to prove that $x(\r{(\theta_1\cap\theta_2)\vee(\theta_1\cap\theta_3)})y$.
Start with $(x\wedge z)\r{\theta_1}(y\wedge z)$, which gives $x\r{\theta_1}((y\wedge z)\vee x)$. Similarly, $(x\wedge y)\r{\theta_2}(z\wedge y)$ and $x\r{\theta_2}((y\wedge z)\vee x)$. This gives $x\r{(\theta_1\cap\theta_2)}((y\wedge z)\vee x)$
I expect a similar manipulation should now give $((y\wedge z)\vee x)\r{\theta_1\cap\theta_3} y$ but I can't see how to manage this. Is this even in the right direction? I suppose something other than $((y\wedge z)\vee x$ could be the middle link, but nothing obviously better comes to mind.