Question

Draw $\Delta \text{ABC}$ and let D be the midpoint of $\overline{\text{AB}}$. Let E be the midpoint of $\overline{\text{CD}}$. Let F be the intersection of $\overrightarrow{\text{AE}}$ and $\overline{\text{BC}}$. Draw$\overline{\text{DG}}$ parallel to$\overline{\text{EF}}$ meeting$\overline{\text{BC}}$ at G. Prove that $\text{BG}=\text{GF}=\text{FC}$ .

Short answer

The diagram is:

The midpoint divides the segment into two congruent segments and Transitive Property proves $\text{BG}=\text{GF}=\text{FC}$.

Step-by-step solution

Step 1. Consider the diagram.

The $\Delta ABC$is:

In $\Delta ABC$, D be the midpoint of $\overline{AB}$. Let E be the midpoint of $\overline{CD}$. Let F be the intersection of $\overrightarrow{AE}$and $\overline{BC}$. Draw $\overline{DG}$parallel to $\overline{EF}$ meeting$\overline{BC}$ at G.

Step 2. State the transitive property.

The transitive property states that if $A=B$ and $B=C$, then $A=C$ .

Step 3. Show the proof.

Statement	Reasons
In $\Delta DGC$, E is the midpoint of $\overline{CD}$and $\overline{DG}\parallel \overline{EF}$.	Given.
F is the midpoint of $\overline{GC}$.	A line that contains the midpoint of the side of a triangle and is parallel to another side passes through the midpoint of the third side.
$GF=FC$	Midpoint divides the segment into two congruent segments.
In $\Delta ABF$, D is the midpoint of $\overline{AB}$and $\overline{DG}\parallel \overline{AF}$	Given.
G is the midpoint of $\overline{BF}$.	A line that contains the midpoint of the side of a triangle and is parallel to another side passes through the midpoint of the third side.
$BG=GF$	Midpoint divides the segment into two congruent segments.
$BG=GF=FC$	Transitive Property.

Step 4. State the conclusion.

Therefore, the statement $BG=GF=FC$is proved by:

The midpoint divides the segment into two congruent segments that is$BG=GF=FC$