The perimeter of a parallelogram $ABCD$ is $30$ $cm$. If $D$ and $M$ which is the midpoint of $\overline{DC}$ lie on the circle with diameter $\overline{AB}$, what is the perimeter of $ABMD$?
We know that $\overline{AB}\parallel \overline{CD}$ since $ABCD$ is a parallelogram, so as $DM\in CD,$ then $\overline{AB}\parallel \overline{DM}$. $M\ne C$ so $\overline{BM}$ cannot be parallel to $\overline{AD}$ which makes the quadrilateral $ABMD$ a trapezoid. Even more, since $A,B,M,D$ lie on $k$, the trapezoid is inscribed, so $\overline{AD}=\overline{BM}$. We have $$AO=BO=\dfrac12AB=\dfrac12CD=DM=CM.$$ What else? Thank you in advance!
Since $\overline{DM}$ is congruent to the radius of the circle, $\triangle DMO$ is equilateral. By symmetry, $\angle AOD\cong\angle BOM$, so those are both $60^\circ$ and the other two triangles inside the circle must be equilateral as well.
Therefore, each of the six line segments that comprise the perimeter of the triangle are congruent, so each has measure $\frac16\cdot30\text{cm}=5\text{cm}$. Since the trapezoid is made of five such congruent segments, its perimeter is $25\text{cm}$.