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.

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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.