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Let's analyze and solve each part of this geometry problem involving triangle \( ABC \), which is a right triangle at \( A \) with \( AB < AC \). The problem provides additional points \( M \), \( D \), \( E \), and \( K \) with specific conditions. We need to prove the following properties for quadrilaterals and segments within this setup.

### Given Information:
1. \( \triangle ABC \) is a right triangle with \( \angle A = 90^\circ \).
2. \( M \) is the midpoint of segment \( BC \).
3. Point \( D \) is chosen on the opposite ray of \( MA \) such that \( MD = MA \).

### To Prove:
1. **(a)** Prove that quadrilateral \( ABDC \) is a rectangle.
2. **(b)** On ray \( AB \), choose point \( E \) such that \( B \) is the midpoint of \( AE \). Prove that quadrilateral \( BEDC \) is a parallelogram.
3. **(c)** If \( EM \) intersects \( BD \) at point \( K \), prove that \( EK = 2 \cdot KM \).

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### Solution Outline:

#### Part (a): Prove \( ABDC \) is a Rectangle
To prove that \( ABDC \) is a rectangle, we need to show that:
   - \( ABDC \) has four right angles.
   - Or alternatively, we could show that opposite sides are parallel and equal in length.

1. Since \( \triangle ABC \) is a right triangle with \( \angle A = 90^\circ \), segment \( AB \) is perpendicular to \( AC \).
2. \( M \) is the midpoint of \( BC \), so \( M \) divides \( BC \) into two equal parts.
3. \( D \) is placed such that \( MD = MA \) and \( D \) lies on the extension of \( MA \), making \( M \) the midpoint of \( AD \) as well.
4. Using symmetry and the fact that both \( MD = MA \) and \( \triangle ABC \) is right, we can conclude that \( ABDC \) forms a rectangle.

#### Part (b): Prove \( BEDC \) is a Parallelogram
To prove \( BEDC \) is a parallelogram, we need to show that opposite sides are parallel and equal in length.

1. Since \( B \) is the midpoint of \( AE \), and \( M \) is the midpoint of \( BC \), by the midpoint theorem, line segments \( BE \) and \( DC \) are parallel and equal in length.
2. Similarly, since \( M \) is the midpoint of both \( AD \) and \( BC \), line segments \( BD \) and \( EC \) are also parallel and equal.
3. With both pairs of opposite sides equal and parallel, \( BEDC \) is a parallelogram.

#### Part (c): Prove \( EK = 2 \cdot KM \)
To prove that \( EK = 2 \cdot KM \), given that \( EM \) intersects \( BD \) at \( K \):

1. Since \( B \) and \( M \) are midpoints, \( EM \) and \( BD \) intersect in such a way that the point \( K \) divides \( EM \) into segments with a ratio of 2:1.
2. By using properties of similar triangles and the midpoint theorem, we can conclude that \( EK = 2 \cdot KM \).

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Would you like further details on any of these steps, or explanations of theorems involved?

### Related Questions
1. How do we prove that a quadrilateral is a rectangle using perpendicular lines?
2. What are the properties of a parallelogram and how do we use them in geometric proofs?
3. How does the midpoint theorem help in proving properties about triangles and quadrilaterals?
4. What is the significance of segment ratios in proving the intersection properties?
5. How can we generalize this setup for other types of triangles?

### Tip:
When dealing with midpoint and parallel line properties in geometry, consider using the midpoint theorem and properties of parallelograms, as they often simplify complex proofs.

Given a right triangle ABC at A (AB < AC), with M as the midpoint of BC and D on the opposite ray of MA such that MD = MA.

(a) Prove that quadrilateral ABDC is a rectangle.
(b) On ray AB, take point E such that B is the midpoint of AE. Prove that quadrilateral BEDC is a parallelogram.
(c) Given that EM intersects BD at K, prove that EK = 2KM.

This problem explores the properties of quadrilaterals and parallelograms within a right triangle setup. Learn how to prove that quadrilateral ABDC is a rectangle and BEDC is a parallelogram, and analyze segment ratios with a step-by-step solution.

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