Soit \(ABCD\) un parallelogramme. 1. Construire \(E\) l'image de \(A\) par la translation qui transforme \(B\) en \(D\). 2. Construire \(F\) l'image de \(B\) par la translation... Soit \(ABCD\) un parallelogramme. 1. Construire \(E\) l'image de \(A\) par la translation qui transforme \(B\) en \(D\). 2. Construire \(F\) l'image de \(B\) par la translation qui transforme \(A\) en \(C\). 3. Montrer que \(E\) est l'image de \(D\) par la translation qui transforme \(C\) en \(D\). 4. Montrer que \(F\) est l'image de \(C\) par la translation qui transforme \(D\) en \(C\). 5. En déduire que : \(\overrightarrow{ED} = \overrightarrow{DC} = \overrightarrow{CF}\) et \(\overrightarrow{EF} = 3\overrightarrow{AB}\) Read more

Steps to Solve

1. Understanding the translations

We are given that $E$ is the image of $A$ by the translation that maps $B$ to $D$. This means $\overrightarrow{AE} = \overrightarrow{BD}$. Also, $F$ is the image of $B$ by the translation that maps $A$ to $C$. This means $\overrightarrow{BF} = \overrightarrow{AC}$.

2. Parallelogram properties

Since ABCD is a parallelogram, $\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD}$ and $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$.

3. Showing E is the image of D by the translation that maps C to D

We want to show that $E$ is the image of $D$ by the translation that transforms $C$ to $D$, i.e., we need to prove $\overrightarrow{DE} = \overrightarrow{CD}$. We know $\overrightarrow{AE} = \overrightarrow{BD}$, so $\overrightarrow{AE} = \overrightarrow{AD} + \overrightarrow{DE}$. Thus, $\overrightarrow{DE} = \overrightarrow{AE} - \overrightarrow{AD}$. Since $\overrightarrow{AE} = \overrightarrow{BD}$, we have $\overrightarrow{DE} = \overrightarrow{BD} - \overrightarrow{AD}$. Also, since $ABCD$ is a parallelogram, $\overrightarrow{BD} = - \overrightarrow{DB}$, and $\overrightarrow{DB} = \overrightarrow{AB} - \overrightarrow{AD}$.

We also have $\overrightarrow{CD} = - \overrightarrow{DC}$, and since $ABCD$ is a parallelogram, $\overrightarrow{DC} = \overrightarrow{AB}$. Therefore, we want to show that $\overrightarrow{DE} = - \overrightarrow{AB}$. Since $\overrightarrow{AE} = \overrightarrow{BD}$, we have $\overrightarrow{OE} - \overrightarrow{OA} = \overrightarrow{OD} - \overrightarrow{OB}$. Thus, $\overrightarrow{OE} = \overrightarrow{OA} + \overrightarrow{OD} - \overrightarrow{OB}$. Now, $\overrightarrow{DE} = \overrightarrow{OE} - \overrightarrow{OD} = \overrightarrow{OA} - \overrightarrow{OB} = \overrightarrow{BA} = - \overrightarrow{AB}$. Thus we have proved $\overrightarrow{DE} = \overrightarrow{CD}$. So, $E$ is the image of $D$ by the translation that transforms $C$ to $D$.

4. Showing F is the image of C by the translation that maps D to C

We want to show that $F$ is the image of $C$ by the translation that transforms $D$ to $C$, i.e., we need to prove $\overrightarrow{CF} = \overrightarrow{DC}$. We know $\overrightarrow{BF} = \overrightarrow{AC}$, so $\overrightarrow{CF} = \overrightarrow{BF} - \overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{BC}$. Since $ABCD$ is a parallelogram, $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$. Thus, $\overrightarrow{CF} = \overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{BC} = \overrightarrow{AB}$. Also, since $ABCD$ is a parallelogram, $\overrightarrow{DC} = \overrightarrow{AB}$. Therefore, $\overrightarrow{CF} = \overrightarrow{DC}$. So, $F$ is the image of $C$ by the translation that transforms $D$ to $C$.

5. Deductions

From step 3, we have $\overrightarrow{DE} = \overrightarrow{CD}$, so $\overrightarrow{ED} = \overrightarrow{DC}$. From step 4, we have $\overrightarrow{CF} = \overrightarrow{DC}$. Therefore, $\overrightarrow{ED} = \overrightarrow{DC} = \overrightarrow{CF}$.

Now let's find $\overrightarrow{EF}$. We have $\overrightarrow{EF} = \overrightarrow{ED} + \overrightarrow{DC} + \overrightarrow{CF}$. Since $\overrightarrow{ED} = \overrightarrow{DC} = \overrightarrow{CF}$, we have $\overrightarrow{EF} = \overrightarrow{DC} + \overrightarrow{DC} + \overrightarrow{DC} = 3\overrightarrow{DC}$. Since $ABCD$ is a parallelogram, $\overrightarrow{DC} = \overrightarrow{AB}$. Therefore, $\overrightarrow{EF} = 3\overrightarrow{AB}$.