Question

question 1-5
in the figure, $overleftrightarrow{ad}paralleloverleftrightarrow{bc}$, $overleftrightarrow{ab}paralleloverleftrightarrow{dc}$, $overleftrightarrow{ab}perpoverleftrightarrow{ad}$, and $angle aedcongangle ade$.
what is the degree measure of $angle dbc$?
$mangle dbc=square^{circ}$

Explanation:

Step1: Identify the shape

Since $\overrightarrow{AD}\parallel\overrightarrow{BC}$, $\overrightarrow{AB}\parallel\overrightarrow{DC}$, and $\overrightarrow{AB}\perp\overrightarrow{AD}$, the quadrilateral $ABCD$ is a rectangle. In a rectangle, each angle is $90^{\circ}$, so $\angle BAD = 90^{\circ}$.

Step2: Analyze $\triangle ABD$

Given $\angle ABD=\angle ADB$. In $\triangle ABD$, since $\angle BAD = 90^{\circ}$ and $\angle ABD+\angle ADB+\angle BAD=180^{\circ}$ (angle - sum property of a triangle), and $\angle ABD=\angle ADB$, we have $2\angle ADB + 90^{\circ}=180^{\circ}$. Solving for $\angle ADB$:
\[

$$\begin{align*} 2\angle ADB&=180^{\circ}- 90^{\circ}\\ 2\angle ADB&=90^{\circ}\\ \angle ADB& = 45^{\circ} \end{align*}$$

\]

Step3: Find $\angle DBC$

In rectangle $ABCD$, $\angle ABC = 90^{\circ}$. Also, $\angle ABC=\angle ABD+\angle DBC$. We know $\angle ABD = 45^{\circ}$, so $\angle DBC=\angle ABC-\angle ABD$.
\[

$$\begin{align*} \angle DBC&=90^{\circ}-45^{\circ}\\ \angle DBC&=45^{\circ} \end{align*}$$

\]

Answer:

$45$