Can you find any of the shown segments? Also, a hint: use the right triangle altitude theorem.
$\displaystyle |AM|=\sqrt{10^2-6^2}=8$
Since $\displaystyle \Delta AMC \sim \Delta BCM$ it follows :
$\displaystyle |BM| : 6 = 6 : 8$
$\displaystyle |BM|=\frac{9}{2} \Rightarrow |AB|=\frac{25}{2}$
Hence :
$\displaystyle |BC|=\sqrt{\left(\frac{25}{2}\right)^2-10^2}=\frac{15}{2}$
therefore perimeter is : $\displaystyle 30$
Here is a slightly different approach:
1. Determine $\displaystyle |\overline{AM}| = 8$ by Pythagorean theorem.
2. Use Pythagorean theorem in triangle ABC:
$\displaystyle (8+\overline{BM})^2-10^2 = (\overline{BC})^2$ and in the small right triangle:
$\displaystyle 6^2+ (\overline{BM})^2 = (\overline{BC})^2$
3. Subtract both equations columnwise and solve for $\displaystyle \overline{BM}$
Of course you are right but you also can use 2 right triangles to prove the altitude theorem:
(Using the labels of the attached diagram):
$\displaystyle CM^2+BM^2=BC^2$
$\displaystyle CM^2+AM^2=AC^2$
Add both equations columnwise:
$\displaystyle 2CM^2 + AM^2 + BM^2 = BC^2+AC^2 = AB^2$
Since $\displaystyle AB = AM + BM$ the right side of the last equation becomes:
$\displaystyle 2CM^2 + AM^2 + BM^2 = (AM + BM)^2 = AM^2 + 2AM \cdot BM + BM^2$
Collect like terms, divide through by 2 and you'll get the equation of the altitude theorem.