Parallelogram and triangles 
Consider the figure where $\square ABCD $ is a parallelogram, $\overline{CK}\perp\overline{AB} $ and $\angle M=90º$. Find:


*

*If $BC=12$, $DM=15$ and $KC=9$. Determine $DC$ and $CM$.

*If $KC= \sqrt{24}$, $AK=\sqrt{18}$, and $KB=\sqrt{8}$. Determine $AD$ and $DM$.


I couldn't get any result in point 1: only the value of $KB$ and that $\angle B \cong \angle MCD$.
In point 2, by pythagorean theorem I found $=\sqrt{24+8}= \sqrt{32}=BC=AD$. I'm sort of stuck in this problem.
 A: For $1)$ Look carefully at the figure and note that $(\triangle DCB)=BC \cdot DM $ Because $\overline{DM}$ is a Height of $\triangle DCB$ with base $\overline{BC}$. Now  $(\triangle DCB)=DC \cdot KC $ Because $\overline{KC}$ is a Height of $\triangle DCB$ with base $\overline{DC}$. 
Hence $BC \cdot DM= DC \cdot KC$, now $DC= \frac{BC \cdot DM}{KC}=\frac{12\cdot15}{9}=20$. 
You can find easily $CM$ using the Pythagorean theorem 
$$CM=\sqrt{DC^2-CM^2}=\sqrt{400-225}=\sqrt{175} \approx 13.22$$
A: Consider the angle $∠DCM$, then $∠BCK=90º-∠DCM$. But also $∠CDM=90º-∠DCM$ so $∠BCK=∠CDM$ and we thus have similar triangles. So $\overline{KC}$ is to $\overline{DM}$ as $\overline{BC}$ is to $\overline{DC}$. Thus $\frac{9}{12}=\frac{15}{\overline{DC}}$ and $\overline{DC}=20$.
Next, use Pythagoras to yield $\overline{CM}$ from $\sqrt{20^2-15^2}≈13.22$
For the second question, the method is very similar, establish a similar triangle relationship and solve. I will leave this for you to try.
A: For point 2:
Consider $\triangle DCB$ with area $a(\triangle DCB)=CB·DM=DC·KC\Rightarrow DM = \frac{DC·KC}{CB} = \frac{(\sqrt{18}+\sqrt{8})(\sqrt{24})}{\sqrt{32}}. $
