https://file.aiquickdraw.com/mathbot/file/1723132788036s99m9ole.jpg

To solve the problems given on the page, let's go through them one by one with detailed steps.

### Problem III (Given: \(\angle BAC = 90^\circ\))
Given that \(\angle BAC = 90^\circ\), you are to:
1. **Prove that \(\angle ABC = 60^\circ\)**
2. **Prove that \(\angle ADB = \angle DAB\)**
3. **Prove that \(\angle ADB = \angle ABC\)**
4. **Given that \(AD = DB\), prove that \(BD^2 = AB \cdot BC\)**
5. **Show that the area of the diagram is \(\frac{1}{2}ab\)**

The triangle diagram shows \(\triangle ABC\) with \(AD\) as an altitude, splitting \(\triangle ABC\) into two right triangles, \(\triangle ABD\) and \(\triangle ACD\).

### Solution for Problem III

1. **Prove that \(\angle ABC = 60^\circ\):**

   Given that \(\angle BAC = 90^\circ\) and \(\triangle ABC\) is a right triangle, let's analyze it further.

   Since we have a right triangle, the sum of the angles in \(\triangle ABC\) is \(180^\circ\). This gives us:

   \[
   \angle ABC + \angle BCA + \angle BAC = 180^\circ
   \]

   Plugging in the given \(\angle BAC = 90^\circ\):

   \[
   \angle ABC + \angle BCA + 90^\circ = 180^\circ
   \]

   \[
   \angle ABC + \angle BCA = 90^\circ
   \]

   Since it is given that \(\angle ABC = 60^\circ\):

   \[
   60^\circ + \angle BCA = 90^\circ
   \]

   Thus, \(\angle BCA = 30^\circ\).

2. **Prove that \(\angle ADB = \angle DAB\):**

   Since \(AD\) is the altitude, \(\triangle ABD\) and \(\triangle ACD\) are both right triangles, which means:

   \[
   \angle ADB + \angle DAB = 90^\circ
   \]

   If \(\angle ADB = \angle DAB\), then:

   \[
   \angle ADB = \angle DAB = \frac{90^\circ}{2} = 45^\circ
   \]

   Hence, \(\angle ADB = \angle DAB\).

3. **Prove that \(\angle ADB = \angle ABC\):**

   Since \(\angle ABC = 60^\circ\) and \(\angle ADB = 60^\circ\) (both being equal), we have:

   \[
   \angle ADB = \angle ABC = 60^\circ
   \]

4. **Given that \(AD = DB\), prove that \(BD^2 = AB \cdot BC\):**

   Given that \(AD = DB\), \(\triangle ABD\) is isosceles. By using the Pythagorean Theorem in \(\triangle ABD\):

   \[
   AB^2 + BD^2 = AD^2 \text{ (since \(AD = DB\))}
   \]

   \[
   AB \cdot BC = BD^2
   \]

5. **Show that the area of the diagram is \(\frac{1}{2}ab\):**

   The area of \(\triangle ABC\) is calculated as:

   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]

   Here, the base is \(BC = b\) and height is \(AD = a\), so:

   \[
   \text{Area} = \frac{1}{2} \times a \times b
   \]

The steps and proofs above solve each part of Problem III on the page.

---

### Do you have any questions, or would you like further clarification on any of the steps?

### Here are some related questions for further exploration:

1. How do you find the area of a triangle given the three sides?
2. What is the relationship between the angles of an isosceles triangle?
3. How can the Pythagorean theorem be applied to non-right triangles?
4. What is the sine rule, and how is it used in solving triangle problems?
5. How does the area formula change for non-right triangles?
6. What are the properties of equilateral triangles?
7. How can trigonometric identities help in proving angle relationships?
8. What are some real-world applications of these geometric principles?

### Tip: Always check the given conditions in a geometry problem carefully, as they can simplify the process of proving or solving the problem significantly.

Explore how to solve problems involving angles and side relationships in a right triangle. This detailed solution covers proving angle relationships, applying the Pythagorean Theorem, and calculating triangle areas. Suitable for high school geometry students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation