Let's break down the information from the diagram and solve the problems step by step.

Given:

1. $BD$ bisects $\angle ABC$.
2. $DE \parallel AB$.
3. $AB \perp BC$.

This forms a triangle $\triangle ABC$ with several important geometric properties due to the bisector and parallel lines. We'll use these facts to find the angles as requested in questions 26 to 30.

Question 26: Find $m \angle DEB$

• Since $DE \parallel AB$ and $AB \perp BC$, $\angle DEB$ is a right angle.
• Therefore, $m \angle DEB = 90^\circ$.

Question 27: Find $m \angle DBE$

• $BD$ bisects $\angle ABC$, meaning it divides $\angle ABC$ into two equal angles.
• Since $AB \perp BC$, we know $\angle ABC = 90^\circ$.
• Thus, $\angle ABD = \angle DBC = 45^\circ$.
• Therefore, $m \angle DBE = 45^\circ$.

Question 28: Find $m \angle EDB$

• Triangle $\triangle DEB$ is a right triangle (from question 26, $m \angle DEB = 90^\circ$).
• Using the triangle sum theorem: $m \angle DEB + m \angle DBE + m \angle EDB = 180^\circ$ Substituting known values: $90^\circ + 45^\circ + m \angle EDB = 180^\circ$ Solving for $m \angle EDB$: $m \angle EDB = 45^\circ$

Question 29: Find $m \angle ABD$

• As found earlier, $BD$ bisects $\angle ABC$, so $m \angle ABD = 45^\circ$.

Question 30: Find an angle congruent to $\angle BAD$

• By the parallel line property, $\angle BAD$ corresponds to $\angle ADE$ because of alternate interior angles (since $AB \parallel DE$).
• Therefore, $\angle ADE \cong \angle BAD$.

Would you like further details or explanations on any step?

Here are 5 related questions to explore further:

1. How do you calculate the length of a segment if one of the angles is known in a right triangle?
2. Can you prove why alternate interior angles are congruent when lines are parallel?
3. What is the relationship between angle bisectors and side ratios in a triangle?
4. How would you find the area of triangle $\triangle ABC$ using the given angles?
5. How would this problem change if $BD$ were not the bisector?

Tip: The sum of angles in any triangle is always $180^\circ$, a fact you can apply in many geometric problems!