Find cd

Extend $AB$ to $E$ such that $\angle DEA=90^\circ$ . Since $\triangle ABC$ is a right isosceles triangle, $\angle A = 45^\circ$ . Let $CD=x$ . Then $AE= DE = \dfrac {16+x}{\sqrt 2}$ and $BE=\dfrac x{\sqrt 2}$ . By Pythagorean theorem, we have:

$\begin{aligned} BE^2 + DE^2 & = BD^2 \\ \frac {x^2}2 + \frac {(16+x)^2}2 & = 17^2 \\ x^2 + x^2 + 32x + 256 & = 578 \\ x^2 + 16x - 161 & = 0 \\ (x-7)(x+23) & = 0 & \small \blue{\text{Since }x > 0} \\ x & = \boxed 7 \end{aligned}$

Let $x=AB=BC$ , then

$16=x\sqrt{2}$

$x=\dfrac{16}{\sqrt{2}}$

Apply Sine Law on $\triangle CDB$

$\dfrac{\sin \angle CDB}{\dfrac{16}{\sqrt{2}}}=\dfrac{\sin 135}{17}$

$\angle CDB = 28.07^\circ$

$angle CBD = 180-135-28.07=16.93^\circ$

Apply Sine Law on $\triangle CDB$ again,

$\dfrac{CD}{\sin 16.93}=\dfrac{17}{\sin 135}$

$\color{#69047E}\large{\boxed{CD = \text{7 units}}}$

$BC = 8\sqrt{2}$ $\; \angle ACB = 45\degree \Rightarrow \angle BCD = 135\degree$

Let $CD = x$

Applying cosine formula in $\triangle BCD$

$17^2 = (8\sqrt{2})^2 + x^2 - 2\cdot(8\sqrt{2})x\text{ cos }135\degree\newline\Rightarrow 289 - 128 = x^2 + 16x\newline\Rightarrow x^2 + 16x - 161 = 0\newline\Rightarrow (x - 7)(x + 23) = 0\newline$ Taking positive value of $x$ .

$CD = x = 7$