Question

type the correct answer in the box. use numerals instead of words. if necessary, use / for the fraction bar.
in $\triangle abc$, $overline{ab}=overline{bc}$ and $mangle abc = 90^{circ}$. $d$ and $e$ are the midpoints of $overline{ab}$ and $overline{bc}$, respectively.
if the length of $overline{ca}$ is 9 units, the length of $overline{de}$ is $square$ units and $mangle cab$ is $square^{circ}$.

Explanation:

Step1: Use the mid - segment theorem

In a triangle, the segment connecting the mid - points of two sides (mid - segment) is parallel to the third side and half its length. In \(\triangle ABC\), \(D\) is the mid - point of \(AB\) and \(E\) is the mid - point of \(BC\). So, \(DE=\frac{1}{2}AC\).
Given \(AC = 9\) units. Then \(DE=\frac{1}{2}\times9 = 4.5\) units.

Step2: Use the properties of an isosceles right - triangle

Since \(AB = BC\) and \(\angle ABC=90^{\circ}\) in \(\triangle ABC\), by the angle - sum property of a triangle (\(\angle A+\angle B+\angle C = 180^{\circ}\)), and \(\angle B = 90^{\circ}\), \(\angle A=\angle C\) (because \(AB = BC\), angles opposite to equal sides are equal).
Let \(\angle CAB=x\) and \(\angle ACB=x\). Then \(x + 90^{\circ}+x=180^{\circ}\).
Simplify the equation: \(2x=180^{\circ}-90^{\circ}=90^{\circ}\).
Solve for \(x\): \(x = 45^{\circ}\).

Answer:

The length of \(\overline{DE}\) is \(4.5\) units and \(m\angle CAB\) is \(45^{\circ}\).