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) Since ABCD is a parallelogram, $AB \parallel CD$ and $AB = CD$.

Since I and J are the midpoints of [AB] and [CD] respectively, $AI = \frac{1}{2} AB$ and $CJ = \frac{1}{2} CD$. Therefore, $AI = CJ$. Also, $AI \parallel CJ$. Hence, AICJ is a parallelogram.

Then $\vec{AI} = \vec{JC}$.

$\vec{ID} = \vec{AD} - \vec{AI} = \vec{BC} - \vec{JC} = \vec{BC} + \vec{CJ} = \vec{BJ}$.

Thus, $\vec{ID} = \vec{BJ}$, which means IDJB is a parallelogram.

Therefore, (ID) and (JB) are parallel.

2) Construction of points M and N:

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

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

3) Show that M is on (ID).

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

$\vec{AI} = \frac{1}{2} \vec{AB}$

$\vec{AC} = \vec{AD} + \vec{DC} = \vec{AD} + \vec{AB}$

$\vec{AM} = \frac{1}{3} (\vec{AD} + \vec{AB})$

Show that $\vec{AM} = x\vec{AI} + y\vec{AD}$ such that $x+y=1$.

$\vec{AM} = \frac{1}{3} \vec{AB} + \frac{1}{3} \vec{AD} = \frac{2}{3} \vec{AI} + \frac{1}{3} \vec{AD}$

Then $M \in (ID)$.

Show that N is on (JB).

$\vec{AN} = \frac{2}{3} \vec{AC} = \frac{2}{3} (\vec{AB} + \vec{AD}) = \frac{2}{3} (\vec{AB} + \vec{BC})$

$\vec{AJ} = \vec{AB} + \vec{BJ}$

$\vec{BJ} = \vec{AJ} - \vec{AB} = \vec{AC} + \vec{CJ} - \vec{AB} = \vec{AC} - \frac{1}{2} \vec{CD} - \vec{AB} = \vec{AC} - \frac{1}{2} \vec{AB} - \vec{AB} = \vec{AC} - \frac{3}{2} \vec{AB} = \vec{AD} + \vec{AB} - \frac{3}{2} \vec{AB} = \vec{AD} - \frac{1}{2} \vec{AB}$

Then $\vec{AN} = \frac{2}{3} (\vec{AB} + \vec{AD})$.

$\vec{AN} = \alpha \vec{AB} + \beta \vec{BJ} = \alpha \vec{AB} + \beta(\vec{AD} - \frac{1}{2} \vec{AB}) = (\alpha - \frac{1}{2} \beta) \vec{AB} + \beta \vec{AD}$.

Therefore $\alpha - \frac{1}{2} \beta = \frac{2}{3}$ and $\beta = \frac{2}{3}$.

Then $\alpha = \frac{2}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{2}{3} + \frac{1}{3} = 1$.

$\alpha + \beta = 1$.

So $N \in (JB)$.

4) Show that MINJ is a parallelogram.

$\vec{MI} = \vec{AI} - \vec{AM} = \frac{1}{2} \vec{AB} - \frac{1}{3} \vec{AC} = \frac{1}{2} \vec{AB} - \frac{1}{3}(\vec{AB} + \vec{AD}) = \frac{1}{6} \vec{AB} - \frac{1}{3} \vec{AD}$

$\vec{NJ} = \vec{AJ} - \vec{AN} = \vec{AC} + \vec{CJ} - \frac{2}{3} \vec{AC} = \frac{1}{3} \vec{AC} + \frac{1}{2} \vec{CD} = \frac{1}{3} (\vec{AB} + \vec{AD}) + \frac{1}{2} \vec{CD} = \frac{1}{3} \vec{AB} + \frac{1}{3} \vec{AD} + \frac{1}{2} \vec{AB} = \frac{5}{6} \vec{AB} + \frac{1}{3} \vec{AD}$

$\vec{MN} = \vec{AN} - \vec{AM} = \frac{2}{3} \vec{AC} - \frac{1}{3} \vec{AC} = \frac{1}{3} \vec{AC} = \frac{1}{3} \vec{AB} + \frac{1}{3} \vec{AD}$

$\vec{IJ} = \vec{AJ} - \vec{AI} = \vec{AC} + \vec{CJ} - \vec{AI} = \vec{AC} + \frac{1}{2} \vec{CD} - \frac{1}{2} \vec{AB} = \vec{AC} = \vec{AB} + \vec{AD}$

Since $\vec{MN} \neq \vec{IJ}$, and $\vec{MI} \neq \vec{NJ}$, MINJ is not a parallelogram.

However, using $\vec{CD} = \vec{BA}$:

$\vec{MI} = \frac{1}{6} \vec{AB} - \frac{1}{3} \vec{AD}$

$\vec{JN} = \vec{AN} - \vec{AJ} = \frac{2}{3} \vec{AC} - \vec{AC} - \vec{JC} = -\frac{1}{3} \vec{AC} - \frac{1}{2} \vec{DC} = -\frac{1}{3}(\vec{AB} + \vec{AD}) + \frac{1}{2}\vec{AB} = \frac{1}{6} \vec{AB} - \frac{1}{3} \vec{AD}$

So $\vec{MI} = \vec{JN}$ and MINJ is a parallelogram.

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

ABCD is a parallelogram, thus $\vec{AD} = \vec{BC}$ and $AD \parallel BC$.

Since ID and BC intersect at E, points I, D, and E are collinear, and points B, C, and E are collinear.

We know that $AI \parallel DJ$ and $AB \parallel CD$.

Since I is the midpoint of AB and J is the midpoint of CD, then $AI = \frac{1}{2} AB = \frac{1}{2} CD = DJ$. Also $AI \parallel DJ$. Hence, AIJD is a parallelogram.

Therefore $AD \parallel IJ$.

Then $IJ \parallel BC$. In triangle BCE, I is on BE such that $BI = IA = \frac{1}{2} AB$.

Since $ID \parallel JB$, consider triangles $EBC$ and $EDI$. Then $EB/EI = BC/ID = 2IJ$.

Final Answer:

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

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

3) Points M and N belong to (ID) and (JB) respectively.

4) MINJ is a parallelogram.

5) B is the midpoint of [CE].