`A B C D E` is a regular pentagon. The bisector of `/_A` of the pentagon meets the side `C D` in `Mdot` Show that `/_A M C=90^0dot`

Answer

To solve the problem, we need to show that the angle \( \angle AMC \) is equal to \( 90^\circ \) in the regular pentagon \( ABCDE \).

### Step-by-Step Solution:

1. **Understanding the Regular Pentagon**:
A regular pentagon has all sides equal and all interior angles equal. The number of sides \( N \) for a pentagon is 5.

2. **Calculating Each Interior Angle**:
The formula for calculating the interior angle of a polygon is:
\[ \text{Interior Angle} = \frac{(N - 2) \times 180^\circ}{N} \]
For a pentagon:
\[ \text{Interior Angle} = \frac{(5 - 2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ \]
Thus, each interior angle of the pentagon \( ABCDE \) is \( 108^\circ \).

3. **Finding the Angle Bisector**:
The bisector of angle \( A \) divides \( \angle A \) into two equal parts. Therefore:
\[ \angle MAB = \angle MAC = \frac{108^\circ}{2} = 54^\circ \]

4. **Analyzing Quadrilateral \( ABMC \)**:
In quadrilateral \( ABMC \), the sum of the interior angles is \( 360^\circ \). Therefore:
\[ \angle MAB + \angle B + \angle C + \angle AMC = 360^\circ \]
Substituting the known angles:
\[ 54^\circ + 108^\circ + 108^\circ + \angle AMC = 360^\circ \]

5. **Calculating \( \angle AMC \)**:
Now, we can simplify the equation:
\[ 54^\circ + 216^\circ + \angle AMC = 360^\circ \]
\[ 270^\circ + \angle AMC = 360^\circ \]
\[ \angle AMC = 360^\circ - 270^\circ = 90^\circ \]

6. **Conclusion**:
Thus, we have shown that:
\[ \angle AMC = 90^\circ \]