Let's work through the questions in the image. Here, BD bisects $\angle ABC$, which means that $\angle ABD = \angle DBC$.

15. Given:

• $\angle ABD = 57^\circ$

Since BD bisects the angle, $\angle DBC = \angle ABD = 57^\circ$.

Now, $\angle ABC = \angle ABD + \angle DBC = 57^\circ + 57^\circ = 114^\circ$.

Answer for 15:

• $\angle DBC = 57^\circ$
• $\angle ABC = 114^\circ$

16. Given:

• $\angle ABD = 70^\circ$

Since BD bisects the angle, $\angle DBC = \angle ABD = 70^\circ$.

Now, $\angle ABC = \angle ABD + \angle DBC = 70^\circ + 70^\circ = 140^\circ$.

Answer for 16:

• $\angle DBC = 70^\circ$
• $\angle ABC = 140^\circ$

17. Given:

• $\angle ABC = 110^\circ$

Since BD bisects the angle, we know that $\angle ABD = \angle DBC$.

Thus, $\angle ABD = \angle DBC = \frac{\angle ABC}{2} = \frac{110^\circ}{2} = 55^\circ$.

Answer for 17:

• $\angle ABD = 55^\circ$
• $\angle DBC = 55^\circ$

Would you like further details or have any other questions?

Here are some related questions to consider:

1. What happens to $\angle ABD$ if $\angle ABC$ increases?
2. How do you prove that BD is an angle bisector using geometry?
3. How would the calculation change if BD did not bisect $\angle ABC$?
4. What is the sum of the interior angles of a triangle, and how does it apply to this diagram?
5. If $\angle ABC = 180^\circ$, what would $\angle ABD$ and $\angle DBC$ be?

Tip: When you know an angle is bisected, you can always divide the larger angle in half to get the two equal smaller angles!