https://r2.erweima.ai/mathbot/image/1726033358548sib7bih0.jpg

Let's analyze the problem from the image.

We are given a triangle \( ABC \), with point \( D \) on side \( AB \), and \( DCB = a^\circ \). The task is to find various angles in terms of \( a \) and show that \( \angle ABC \) is a right angle.

### Part (a)
We are asked to find:

#### i) \( \angle DBC \)
- \( D \) is on \( AB \), and the line segments \( AD \) and \( BD \) are marked as equal, so \( \triangle ABD \) is isosceles.
- In an isosceles triangle, the base angles are equal. Therefore, \( \angle DBC = \angle ADB \).
- Let’s call the unknown angle \( x \). The sum of angles in triangle \( BCD \) is \( 180^\circ \), so:
  \[
  \angle DBC + \angle DCB + \angle BDC = 180^\circ
  \]
  Substituting the known angle \( \angle DCB = a^\circ \):
  \[
  x + a^\circ + x = 180^\circ
  \]
  \[
  2x + a^\circ = 180^\circ
  \]
  \[
  2x = 180^\circ - a^\circ
  \]
  \[
  x = 90^\circ - \frac{a^\circ}{2}
  \]
  Therefore, \( \angle DBC = 90^\circ - \frac{a^\circ}{2} \).

#### ii) \( \angle ADB \)
Since \( \triangle ABD \) is isosceles, and the base angles are equal, \( \angle ADB \) is also \( 90^\circ - \frac{a^\circ}{2} \).

#### iii) \( \angle DBA \)
Using the triangle sum property for \( \triangle ABD \), we know:
\[
\angle ADB + \angle DBA + \angle DAB = 180^\circ
\]
Since \( \angle ADB = 90^\circ - \frac{a^\circ}{2} \) and \( \angle DBA = 90^\circ - \frac{a^\circ}{2} \) from the isosceles property, we can solve for \( \angle DAB \):
\[
2 \left( 90^\circ - \frac{a^\circ}{2} \right) + \angle DAB = 180^\circ
\]
Simplifying:
\[
180^\circ - a^\circ + \angle DAB = 180^\circ
\]
\[
\angle DAB = a^\circ
\]
Therefore, \( \angle DBA = a^\circ \).

### Part (b)
Show that \( \angle ABC \) is a right angle.

- In \( \triangle ABC \), we now have \( \angle DBC = 90^\circ - \frac{a^\circ}{2} \) from part (i).
- Since \( \angle ABC \) and \( \angle DBC \) form a straight line on segment \( BC \), their sum must be \( 180^\circ \).
- Thus, \( \angle ABC + (90^\circ - \frac{a^\circ}{2}) = 180^\circ \), which simplifies to:
  \[
  \angle ABC = 90^\circ
  \]
Therefore, \( \angle ABC \) is indeed a right angle.

---

Would you like further details or have any questions? Here are 5 related questions:

1. What is the significance of the isosceles triangle properties in this problem?
2. How do we use the triangle sum property to find unknown angles?
3. Can we apply this method to find unknown angles in other triangle types?
4. How would the problem change if \( \angle DCB \) was greater than \( 90^\circ \)?
5. What are the key steps to proving that an angle is a right angle?

**Tip:** Remember, the sum of angles in any triangle is always \( 180^\circ \), a crucial property for solving geometry problems.

In the given triangle ABC, point D is on side AB, and the angle DCB is given as a°. You are asked to find angles ∠DBC, ∠ADB, and ∠DBA in terms of a, and show that ∠ABC is a right angle.

In this geometry problem, we explore the properties of an isosceles triangle and use angle sum theorems to find angles ∠DBC, ∠ADB, and ∠DBA in terms of a°, and prove that ∠ABC is a right angle. This solution involves the application of isosceles triangle properties and angle sum rules.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation