In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The lines segments AF and BF meet the line segments ED and EC at points G and H respectively. Prove that :
triangle HEB and FHC are congruent.

Answer

To prove that triangle HEB is congruent to triangle FHC, we will use the properties of a parallelogram and the congruence criteria for triangles. Let's break down the proof step by step.

### Step 1: Identify the Given Information
We have a parallelogram ABCD with E and F as the midpoints of sides AB and CD respectively. The line segments AF and BF intersect line segments ED and EC at points G and H respectively.

### Step 2: Draw the Diagram
Draw the parallelogram ABCD. Mark points E and F as the midpoints of sides AB and CD. Draw line segments AF and BF, and mark their intersections with ED and EC as points G and H.

### Step 3: Establish Corresponding Sides
Since E and F are midpoints:
- \( EB = \frac{1}{2} AB \)
- \( FC = \frac{1}{2} CD \)

Since ABCD is a parallelogram, we know that \( AB = CD \). Therefore:
\[ EB = FC \]

### Step 4: Establish Corresponding Angles
- Angle \( FCH \) is equal to angle \( HEB \) because they are alternate interior angles formed by the transversal line EF cutting through the parallel lines AB and CD.
\[ \angle FCH = \angle HEB \]

- Angle \( EHC \) is equal to angle \( EHB \) because they are vertically opposite angles.
\[ \angle EHC = \angle EHB \]

### Step 5: Apply the Congruence Criteria
We have established:
1. \( EB = FC \) (sides)
2. \( \angle HEB = \angle FCH \) (alternate interior angles)
3. \( \angle EHC = \angle EHB \) (vertically opposite angles)

By the Angle-Side-Angle (ASA) congruence criterion, we can conclude that:
\[ \triangle HEB \cong \triangle FHC \]

### Conclusion
Thus, we have proved that triangle HEB is congruent to triangle FHC.

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