in $ \triangle abc, \overline{ab}=\overline{bc} $ and $ m\angle 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 units and $ m\angle cab $ is $ {}^{\circ} $.

Question

in $ \triangle abc, \overline{ab}=\overline{bc} $ and $ m\angle 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 units and $ m\angle cab $ is $ {}^{\circ} $.

Accepted Answer

# Explanation:
## Step1: Use the mid - segment theorem
The mid - segment theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. In $\triangle ABC$, $D$ is the midpoint of $AB$ and $E$ is the midpoint of $BC$, so $DE=\frac{1}{2}AC$. Given $AC = 9$ units, then $DE=\frac{9}{2}=4.5$ units.
## Step2: Use the properties of a right - isosceles triangle
Since $AB = BC$ and $\angle ABC=90^{\circ}$ in $\triangle ABC$, by the angle - sum property of a triangle ($\angle ABC+\angle BAC+\angle BCA = 180^{\circ}$) and $\angle BAC=\angle BCA$ (because $AB = BC$ in $\triangle ABC$), we have $90^{\circ}+2\angle CAB=180^{\circ}$. Solving for $\angle CAB$:
$$
\begin{align*}
2\angle CAB&=180^{\circ}- 90^{\circ}\
2\angle CAB&=90^{\circ}\
\angle CAB& = 45^{\circ}
\end{align*}
$$

# Answer:
The length of $\overline{DE}$ is $4.5$ units and $m\angle CAB$ is $45^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the mid - segment theorem

Step2: Use the properties of a right - isosceles triangle

Answer: