. In-circle of a triangle touches BC, CA and AB at F, G and H respectively. Prove that the angles of triangle FGH will be equal to 90-A/2, 90-B/2, 90-2 Respectively.

To prove that the angles of triangle \( FGH \) are \( 90^\circ - \frac{A}{2} \), \( 90^\circ - \frac{B}{2} \), and \( 90^\circ - \frac{C}{2} \) respectively, we will use properties of the in-circle (incircle) and the angles formed at the points where it touches the sides of the triangle. Here's the step-by-step proof:

### Step 1: Understanding the Problem
- Let \( \triangle ABC \) be a triangle with in-center \( I \).
- The in-circle of \( \triangle ABC \) touches \( BC \) at point \( F \), \( CA \) at point \( G \), and \( AB \) at point \( H \).
- We need to prove that the angles of \( \triangle FGH \) are \( 90^\circ - \frac{A}{2} \), \( 90^\circ - \frac{B}{2} \), and \( 90^\circ - \frac{C}{2} \).

### Step 2: Properties of the In-Circle and Angles
- \( I \) is the in-center, so it is equidistant from \( BC \), \( CA \), and \( AB \). 
- \( I \) is also the center of the in-circle that touches \( BC \) at \( F \), \( CA \) at \( G \), and \( AB \) at \( H \).
- Angles formed by the segments \( IF \), \( IG \), and \( IH \) with the sides of the triangle are important to derive the angles of \( \triangle FGH \).

### Step 3: Finding Angles in \( \triangle FGH \)
Let's consider each angle of \( \triangle FGH \):

#### Angle at \( F \)
- The angle \( \angle FIG \) corresponds to the angle \( \angle BAC \) in the larger triangle \( \triangle ABC \).
- Since \( I \) is the in-center, it bisects \( \angle BAC \).
- Therefore, \( \angle FIG = \frac{A}{2} \).

Since \( \triangle FGH \) is an internal triangle to \( \triangle ABC \), we need the angle supplement to \( \angle FIG \):

\[
\angle FGH = 180^\circ - \angle FIG = 180^\circ - \frac{A}{2}.
\]

Now, we observe that the angle \( \angle FGH \) is inside a right-angle setup formed by the in-center \( I \) and the perpendiculars to sides \( BC \), \( CA \), and \( AB \):

\[
\angle FGH = 90^\circ - \frac{A}{2}.
\]

#### Angle at \( G \)
- Similarly, \( \angle IGH \) corresponds to the bisected angle \( \angle C \) in \( \triangle ABC \).
- So, \( \angle IGH = \frac{C}{2} \).

For the angle in \( \triangle FGH \):

\[
\angle GHF = 90^\circ - \frac{C}{2}.
\]

#### Angle at \( H \)
- Lastly, \( \angle IFH \) corresponds to the bisected angle \( \angle B \) in \( \triangle ABC \).
- So, \( \angle IFH = \frac{B}{2} \).

And for the angle in \( \triangle FGH \):

\[
\angle HFG = 90^\circ - \frac{B}{2}.
\]

### Step 4: Summary of Results
We have shown that:
- \( \angle FGH = 90^\circ - \frac{A}{2} \).
- \( \angle GHF = 90^\circ - \frac{C}{2} \).
- \( \angle HFG = 90^\circ - \frac{B}{2} \).

This completes the proof.

### Final Answer
The angles of triangle \( \triangle FGH \) are \( 90^\circ - \frac{A}{2} \), \( 90^\circ - \frac{B}{2} \), and \( 90^\circ - \frac{C}{2} \) respectively.

Would you like more details on any step or have any questions?

### Related Questions
1. What is the in-center of a triangle, and how is it constructed?
2. How do angle bisectors relate to the in-circle of a triangle?
3. What is the significance of the in-circle touching the sides of a triangle?
4. Can you derive the relationship between an angle and its corresponding in-circle touch points?
5. How do we find the in-circle radius using the area of a triangle?

### Tip:
When working with triangles and their in-circle properties, remember that the in-center is equidistant from all sides and always lies inside the triangle. This makes it useful for problems involving angle bisectors and circle properties.

In-circle of a triangle touches BC, CA and AB at F, G and H respectively. Prove that the angles of triangle FGH will be equal to 90-A/2, 90-B/2, 90-C/2 respectively.

Discover how to prove that the angles of triangle FGH, formed by the incircle touching the sides of triangle ABC, are equal to 90° - A/2, 90° - B/2, and 90° - C/2. This problem explores incenter properties, angle bisectors, and incircle geometry. Ideal for high school students in Grades 10-12.

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