https://file.aiquickdraw.com/mathbot/file/1721959216748koqj6nsv.png

$$ {{\displaylines15\text{. (}5分)\,如图,\,点C,\,点E分别为AB,\,CD上一点,\,已知AB//EF,\,且CG平分\angle BCD.\,若\angle DEF=\frac{3\pi}{5},\,则\angle BCG\\ =_{--------.}\left(结果保留\pi\right)}} $$

To solve the problem, we need to analyze the given geometric configuration and apply relevant geometric properties and theorems. Let's proceed step-by-step:

1. **Given Information**:
    - $AB \parallel EF$
    - $CG$ bisects $\angle BCD$
    - $\angle DEF = \frac{3\pi}{5}$

2. **Objective**:
    - Find $\angle BCG$ and express the result in terms of $\pi$.

### Step-by-Step Solution:

1. **Parallel Lines and Transversal**:
    Since $AB \parallel EF$, and $CD$ is a transversal, we have:
    \[
    \angle ACD = \angle EFD
    \]
    This is due to corresponding angles being equal when a transversal intersects two parallel lines.

2. **Bisector Property**:
    Since $CG$ bisects $\angle BCD$, we have:
    \[
    \angle BCG = \angle GCD
    \]
    Let $\angle BCG = \theta$. Therefore, $\angle GCD = \theta$.

3. **Angles Sum Around Point**:
    The angles around point $D$ add up to $2\pi$. We can write:
    \[
    \angle BCD = \angle BCG + \angle GCD = 2\theta
    \]

4. **Angle DEF Calculation**:
    Given $\angle DEF = \frac{3\pi}{5}$, we need to find the remaining angle around point $E$. Since $E$ is a point on $CD$ and $AB \parallel EF$, we need to consider the supplementary angles around $E$:
    \[
    \angle DEC + \angle DEF = \pi
    \]
    \[
    \angle DEC = \pi - \frac{3\pi}{5} = \frac{2\pi}{5}
    \]

5. **Sum of Angles at Point D**:
    Considering the angles around point $D$:
    \[
    \angle BCD + \angle DEF = 2\theta + \frac{3\pi}{5}
    \]
    But we also need to consider $\angle EFD$:
    \[
    \angle EFD = \angle ACD = \frac{2\pi}{5} \, \text{(from step 4)}
    \]

6. **Angle Sum Around D**:
    The total angle sum around point $D$ should be:
    \[
    2\theta + \frac{2\pi}{5} + \frac{3\pi}{5} = 2\pi
    \]
    \[
    2\theta + \pi = 2\pi
    \]
    \[
    2\theta = \pi
    \]
    \[
    \theta = \frac{\pi}{2}
    \]

### Final Answer:
Thus, $\angle BCG = \frac{\pi}{2}$.

Would you like more details or have any questions?

Here are some questions you might ask:
1. How do you determine the angles around a point?
2. Can you explain the property of angle bisectors in more detail?
3. Why do the angles around a point sum up to $2\pi$?
4. How does the parallel line property help in solving geometric problems?
5. What are the key theorems used in solving this problem?

**Tip:** Always check the relationship between angles and lines, especially when dealing with parallel lines and transversals, as they often provide helpful angle properties.

Explore a geometry problem involving parallel lines and an angle bisector. Learn how to find angle BCG given specific angle measures and geometric properties. Suitable for high school students studying geometry.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation