Triangle 1

We note that $\triangle ABE$ is similar to $\triangle AEC$ , therefore $\dfrac {AC}{AE}=\dfrac {AE}{AB}=\dfrac {12}8 =\dfrac 32$ $\implies AC = \dfrac 32 AE = \dfrac 32 \times 12=18$ . And $BC =AC-AB =18-8=10$ .

Again $\triangle ABE$ is similar to $\triangle BDC$ , therefore $\dfrac {BC}{BD}=\dfrac {AE}{AB} =\dfrac 32$ $\implies BD = \dfrac 23 BC = \dfrac 23 \times 10=\boxed{\dfrac {20}3}$ .

$\cos A=\dfrac{8}{12}=\dfrac{12}{8+BC}$ $\implies$ $8(8+BC)=12(12)$ $\implies$ $64+8BC=144$ $\implies$ $BC=10$

$EC=\sqrt{18^2-12^2}=\sqrt{180}=6\sqrt{5}$

$BE=\sqrt{12^2-8^2}=\sqrt{80}=4\sqrt{5}$

$\sin \angle BEC = \dfrac{BC}{EC}=\dfrac{BD}{BE}$ $\implies$ $\dfrac{10}{6\sqrt{5}}=\dfrac{BD}{4\sqrt{5}}$ $\implies$ $BD=\dfrac{40\sqrt{5}}{6\sqrt{5}}=\boxed{\dfrac{20}{3}}$

Consider the diagram.

$BE=\sqrt{12^2-8^2}=4\sqrt{5}$

By similar triangles, we have

$\dfrac{x}{4\sqrt{5}}=\dfrac{4\sqrt{5}}{12}$

$x=\dfrac{15(5)}{12}=\dfrac{80}{12}=$ $\boxed{\dfrac{20}{3}}$