#### Answer

$AD = \sqrt{6}+\sqrt{2}$

#### Work Step by Step

We can use the Pythagorean theorem to find the length of $AD$:
$(AD)^2 = (AC)^2+(DC)^2$
$AD = \sqrt{(AC)^2+(DC)^2}$
$AD = \sqrt{(1)^2+(2+\sqrt{3})^2}$
$AD = \sqrt{1+(4+4\sqrt{3}+3)}$
$AD = \sqrt{8+4\sqrt{3}}$
$AD = \sqrt{6+2\sqrt{12}+2}$
$AD = \sqrt{(\sqrt{6}+\sqrt{2})~(\sqrt{6}+\sqrt{2})}$
$AD = \sqrt{6}+\sqrt{2}$