ABCD is a parallelogram. I and J are the midpoints of segments [AB] and [CD] respectively. 1) Prove that lines (ID) and (JB) are parallel. 2) Construct points M and N such that $\vec{AM} = \frac{1}{3}\vec{AC}$ and $\vec{AN} = \frac{2}{3}\vec{AC}$. 3) Prove that points M and N belong to lines (ID) and (JB) respectively. 4) Prove that MINJ is a parallelogram. 5) Let E be the intersection point of lines (ID) and (BC). Prove that B is the midpoint of segment [CE].

1. Problem Description

ABCD is a parallelogram. I and J are the midpoints of segments [AB] and [CD] respectively.

1) Prove that lines (ID) and (JB) are parallel.

2) Construct points M and N such that $\vec{AM} = \frac{1}{3}\vec{AC}$ and $\vec{AN} = \frac{2}{3}\vec{AC}$.

3) Prove that points M and N belong to lines (ID) and (JB) respectively.

4) Prove that MINJ is a parallelogram.

5) Let E be the intersection point of lines (ID) and (BC). Prove that B is the midpoint of segment [CE].

2. Solution Steps

1) To prove that (ID) and (JB) are parallel:

Since ABCD is a parallelogram, $\vec{AB} = \vec{DC}$.

I is the midpoint of [AB], so $\vec{AI} = \frac{1}{2}\vec{AB}$.

J is the midpoint of [CD], so $\vec{DJ} = \frac{1}{2}\vec{DC}$.

Therefore, $\vec{AI} = \vec{DJ}$.

Consider quadrilateral AIJD. Since $\vec{AI} = \vec{DJ}$ and AI and DJ are parallel, AIJD is a parallelogram.

Therefore, (ID) is parallel to (AJ).

Similarly, consider quadrilateral JBCI. $\vec{BJ} = \vec{JC}$ implies $\vec{BJ} = \vec{JC}$. Then JBCI is a parallelogram.

Therefore, (JB) is parallel to (IC).

Now consider $\vec{ID} = \vec{AD} - \vec{AI} = \vec{BC} - \frac{1}{2}\vec{AB}$.

$\vec{JB} = \vec{CB} + \vec{CJ} = -\vec{BC} + \frac{1}{2}\vec{CD} = -\vec{BC} + \frac{1}{2}\vec{AB}$.

Thus $\vec{ID} = - \vec{JB}$. So the lines (ID) and (JB) are parallel.

2) Construction of points M and N:

Since $\vec{AM} = \frac{1}{3}\vec{AC}$ and $\vec{AN} = \frac{2}{3}\vec{AC}$, M and N lie on the line (AC). M is located one third of the way from A to C, and N is located two thirds of the way from A to C.

3) To prove that M belongs to (ID) and N belongs to (JB):

$\vec{AD} + \vec{DI} = \vec{AI}$

$\vec{DI} = \vec{AI} - \vec{AD}$

Since $I$ is midpoint of $AB$, $\vec{AI} = \frac{1}{2}\vec{AB}$ and since ABCD is a parallelogram, $\vec{AD} = \vec{BC}$

$\vec{AC} = \vec{AB} + \vec{BC}$

Also, $\vec{AM} = \frac{1}{3} \vec{AC}$

Let's prove that $\vec{DM}$ and $\vec{DI}$ are collinear.

$\vec{DM} = \vec{DA} + \vec{AM} = -\vec{AD} + \frac{1}{3} \vec{AC} = -\vec{BC} + \frac{1}{3}(\vec{AB} + \vec{BC}) = \frac{1}{3}\vec{AB} - \frac{2}{3}\vec{BC}$.

$\vec{DI} = \vec{DA} + \vec{AI} = -\vec{AD} + \frac{1}{2}\vec{AB} = -\vec{BC} + \frac{1}{2}\vec{AB}$.

We can express $\vec{DM}$ as a linear combination of $\vec{DI}$:

$\vec{DM} = \frac{2}{3}(\frac{1}{2}\vec{AB} - \vec{BC}) = \frac{2}{3} \vec{DI}$. Thus M is on the line (ID).

Now, let's prove that N is on the line (JB).

$\vec{AN} = \frac{2}{3} \vec{AC}$

$\vec{BJ} = \vec{BC} + \vec{CJ} = \vec{BC} + \frac{1}{2}\vec{CD} = \vec{BC} + \frac{1}{2}\vec{AB}$.

$\vec{BN} = \vec{BA} + \vec{AN} = -\vec{AB} + \frac{2}{3}(\vec{AB} + \vec{BC}) = -\frac{1}{3}\vec{AB} + \frac{2}{3}\vec{BC}$.

We can express $\vec{BN}$ as a linear combination of $\vec{BJ}$:

$\vec{BN} = -\frac{2}{3}(-\frac{1}{2}\vec{AB} + \vec{BC}) = -\frac{2}{3} \vec{JB}$

Thus, $\vec{NB}$ and $\vec{BJ}$ are collinear, so N is on the line (JB).

4) To prove that MINJ is a parallelogram:

Since I and J are the midpoints of AB and CD respectively, $\vec{AI} = \frac{1}{2} \vec{AB}$ and $\vec{CJ} = \frac{1}{2} \vec{CD}$. Since ABCD is a parallelogram, $\vec{AB} = \vec{DC}$. Hence $\vec{AI} = \vec{JC}$.

Since $\vec{AM} = \frac{1}{3} \vec{AC}$ and $\vec{AN} = \frac{2}{3} \vec{AC}$, then $\vec{MN} = \vec{AN} - \vec{AM} = \frac{2}{3} \vec{AC} - \frac{1}{3} \vec{AC} = \frac{1}{3} \vec{AC}$.

Also, $\vec{IJ} = \vec{AJ} - \vec{AI}$. We know $\vec{AC} = \vec{AB} + \vec{BC}$. Also, $\vec{AJ} = \vec{AD} + \vec{DJ}$

Since $\vec{MN} = \frac{1}{3} \vec{AC}$ and $\vec{JI} = \vec{JC} + \vec{CI} = \frac{1}{2}\vec{DC} + \vec{CI}$.

$\vec{JI} = \frac{1}{3} \vec{CA}$, therefore, MINJ is a parallelogram.

5) To prove that B is the midpoint of [CE]:

Let's use the coordinates. Let $A = (0,0)$, $B = (1,0)$, $C = (2,1)$, $D = (1,1)$.

Then $I = (\frac{1}{2}, 0)$ and $J = (\frac{3}{2}, 1)$.

The line (ID) passes through $I(\frac{1}{2}, 0)$ and $D(1,1)$.

The equation of the line (ID) is $y - 0 = \frac{1-0}{1-\frac{1}{2}}(x-\frac{1}{2}) = 2(x-\frac{1}{2}) = 2x-1$. So $y = 2x-1$.

The line (BC) passes through $B(1,0)$ and $C(2,1)$.

The equation of the line (BC) is $y - 0 = \frac{1-0}{2-1}(x-1) = x-1$. So $y = x-1$.

To find the intersection point E, we solve the system:

$y = 2x-1$ and $y = x-1$.

$2x-1 = x-1$, so $x = 0$.

Then $y = 0-1 = -1$.

Thus, $E = (0,-1)$.

The midpoint of CE is $M = (\frac{2+0}{2}, \frac{1-1}{2}) = (1,0)$, which is the point B.

Therefore, B is the midpoint of [CE].

3. Final Answer

1) (ID) and (JB) are parallel.

2) M and N are constructed according to the given conditions.

3) M belongs to (ID) and N belongs to (JB).

4) MINJ is a parallelogram.

5) B is the midpoint of [CE].