Question

Quad WXYZ must be a special figure to meet the conditions stated. Write the best name for that special quadrilateral.

M is the midpoint of hypotenuse $AM$and$m\angle ACM$

Short answer

$AM=17,m\angle ACM=67$

Step-by-step solution

Step 1. Check the figure.

Consider the figure.

Step 2. Step description.

Consider the triangle $\Delta ACB$.

Now, $\overline{AB}$is the hypotenuse of the right triangle $\Delta ACB$ and M is its midpoint.

So, by the theorem, the midpoint M of the hypotenuse $\overline{AB}$of a right triangle $\Delta ACB$is equidistant from the three vertices.

Thus,$BM=AM=CM=17$

Step 3. Step description.

Now, consider the triangle $\Delta MCB$, here $\overline{BM}\cong \overline{AM}$, so by theorem , the angles $\angle MCB$and $\angle MBC$are congruent.

That is, $m\angle MCB=m\angle MBC=23$.

As $m\angle ACB=90$, so

$\begin{array}{c}m\angle ACB=90\\ m\angle ACM+m\angle MCB=90\\ m\angle ACM+23=90\end{array}$

$\begin{array}{c}m\angle ACM=90-23\\ m\angle ACM=67\end{array}$

Therefore, $AM=17,m\angle ACM=67$.