Chapter 5.3, Problem 24WE

To determine

To prove: that $BG=GD=FC$ with the help of given information.

Answer to Problem 24WE

The Midpoint divides the segment into two congruent segments and Transitive Property i.e. $BG=GF=FC$

Explanation of Solution

Given information:

Given: $\text{Δ}ABC,D$ is the midpoint of side $\overline{AB}$ and $E$ is the midpoint of $\overline{CD}$

And $F$ is the point of intersection of $\overrightarrow{AE}$ and $\overline{BC}$

And $\overline{DG}$ is parallel to $\overline{EF}$ meeting $\overline{BC}$ at $G$

Calculation:

Consider the statement above and associated figure.

The above statement can be proved as follows:

Statement	Reasons
In $△DGC$ $E$ is the midpoint of $\overline{CD}\text{ and }\overline{DG}\parallel \overline{EF}$	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
$\text{In Δ}ABF$ $D$ is the midpoint of $\overline{AB}$ and $DG\parallel \overline{AF}$	Given
$\text{G}$ is the midpoint of $BF$	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

Hence, the above statement is proved by:

The Midpoint divides the segment into two congruent segments and Transitive Property i.e.

  $BG=GF=FC$