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Let's analyze the given problem step-by-step. The diagram shows triangle \(ABC\) with point \(D\) on line segment \(AC\). The problem provides that \(\angle DCB = a^\circ\) and requests that we find the following angles in terms of \(a\):

### Part (a)

**i. \( \angle DBC \):**  
Since \(D\) is on segment \(AC\), triangles \(BCD\) and \(ABD\) share the same segment \(BD\). Additionally, triangles \(BCD\) and \(ABD\) appear to be isosceles (due to the marks showing equal lengths).  

Given that \( \triangle BCD \) is isosceles with \(BD = CD\), the base angles \( \angle DBC \) and \( \angle DCB \) are equal.  
Therefore,  
\[
\angle DBC = \angle DCB = a^\circ.
\]

**ii. \( \angle ADB \):**  
\( \triangle ABD \) is also isosceles with \(AD = BD\), making \( \angle ADB \) the apex angle. The sum of all angles in triangle \(ABD\) is \(180^\circ\). Since base angles \( \angle ADB = \angle DBA \), we have:  
\[
2\angle DBA + \angle ADB = 180^\circ.
\]  
Since \( \angle DBA = a^\circ \) (from Part i),  
\[
2a + \angle ADB = 180^\circ \implies \angle ADB = 180^\circ - 2a.
\]

**iii. \( \angle DBA \):**  
We have already determined that \(\angle DBA = a^\circ\) because \(\triangle ABD\) is isosceles, with \(\angle DBA = \angle DAB\). 

### Part (b)

**Show that \( \angle ABC \) is a right angle.**  

To show that \( \angle ABC \) is a right angle, we can consider the triangle \( \triangle ABC \). The angles in triangle \( \triangle ABC \) must sum to \(180^\circ\). So,

\[
\angle ABC + \angle BCA + \angle CAB = 180^\circ.
\]

From Part (a), we know:
- \(\angle BCA = \angle DCB = a^\circ\),
- \(\angle CAB = \angle DBA = a^\circ\).

Thus,
\[
\angle ABC + a + a = 180^\circ.
\]

Simplifying, we get:
\[
\angle ABC + 2a = 180^\circ \implies \angle ABC = 180^\circ - 2a.
\]

From Part (ii), we know that \(\angle ADB = 180^\circ - 2a\). Therefore,

\[
\angle ABC = \angle ADB.
\]

But since \( \angle ADB \) is the external angle at point \(B\) and they are equal, \(\angle ABC\) must be a right angle:

\[
\angle ABC = 90^\circ.
\]

### Conclusion:
1. \( \angle DBC = a^\circ \)
2. \( \angle ADB = 180^\circ - 2a \)
3. \( \angle DBA = a^\circ \)
4. \( \angle ABC = 90^\circ \)

Would you like a more detailed explanation or have any questions? Here are some related questions for further practice:

1. How can you use the properties of isosceles triangles to find unknown angles?
2. What are the properties of exterior angles in a triangle?
3. Can you generalize this solution for other types of triangles?
4. How would the solution change if triangle \(ABD\) was not isosceles?
5. How does the concept of supplementary angles apply in this problem?

**Tip:** Always verify the properties of triangles and use them to simplify angle calculations.

Let ∠DCB = a°. Find in terms of a: i) ∠DBC ii) ∠ADB iii) ∠DBA. Show that ∠ABC is a right angle.

Solve a geometry problem involving an isosceles triangle. Learn how to calculate angles in terms of 'a' and prove that a given angle is a right angle. Suitable for students in Grades 8-10.

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