$$ \overrightarrow{F H} $$ bisects $$ \angle E F G $$. Find the indicated angle measures. $$ m \angle E F H=63^{\circ} $$. Find $$ m \angle G F H $$ and $$ m \angle E F G $$. $$ m \angle G F H=\square^{\circ} $$ $$ m \angle E F G=\square^{\circ} $$

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To solve the problem, let's analyze the given information step by step.

### Given:
1. **Ray $$ \overrightarrow{FH} $$** bisects the angle $$ \angle EFG $$. This means that $$ \overrightarrow{FH} $$ divides $$ \angle EFG $$ into two equal parts.
2. **Measure of $$ \angle EFH = 63^\circ $$**.

### Find:
1. **Measure of $$ \angle GFH $$**
2. **Measure of $$ \angle EFG $$**

### Solution:

1. **Understanding the Angle Bisector:**
   
   Since $$ \overrightarrow{FH} $$ bisects $$ \angle EFG $$, it divides $$ \angle EFG $$ into two congruent angles:
   $$
   \angle EFH = \angle GFH
   $$
   
2. **Given $$ \angle EFH = 63^\circ $$:**
   
   Therefore,
   $$
   \angle GFH = 63^\circ
   $$
   
3. **Calculating $$ \angle EFG $$:**
   
   Since $$ \angle EFG $$ is the sum of $$ \angle EFH $$ and $$ \angle GFH $$:
   $$
   \angle EFG = \angle EFH + \angle GFH = 63^\circ + 63^\circ = 126^\circ
   $$

### **Final Answers:**
- $$ m \angle GFH = 63^\circ $$
- $$ m \angle EFG = 126^\circ $$
$$ m \angle GFH = 63^\circ $$ and $$ m \angle EFG = 126^\circ $$.

\( \overrightarrow{F H} \) bisects \( \angle E F G \). Find the indicated angle measures. \( m \angle E F H=63^{\circ} \). Find \( m \angle G F H \) and \( m \angle E F G \). \( m \angle G F H=\square^{\circ} \) \( m \angle E F G=\square^{\circ} \)

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