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.