Chapter 4.7, Problem 27WE

a.

To determine

To prove :  That $\overline{GQ}\cong \overline{GP}$ in paragraph form.

Explanation of Solution

Given information :  Consider a scalene $\Delta EFG$ . $\overline{EH}$ is a median and $P$ is a point on $\overrightarrow{EH}$ such that $\overline{EH}\cong \overline{HP}$ . Also, $\overline{FJ}$ is a median and $Q$ is a point on $\overrightarrow{FJ}$ such that $\overline{FJ}\cong \overline{JQ}$ . It is to be proved in paragraph form that $\overline{GQ}\cong \overline{GP}$ .

  

Formula used : The SAS postulate states that for two triangles to be congruent two sides and included angle of one triangle should be congruent to two sides and included angle of another triangle.

Corresponding parts of congruent triangles are congruent. 

Proof :  Consider $\Delta EJF$ and $\Delta GJQ$ , as shown in the figure below,

  

Given that $\overline{JF}\cong \overline{JQ}$ , also given that $\overline{FJ}$ is a median and a median from $F$ to the opposite $\overline{EG}$ bisects $\overline{EG}$ hence, $\overline{EJ}\cong \overline{GJ}$ . Vertically opposite angles are equal so $\angle EJF\cong \angle GJQ$ . From these properties it can be concluded by SAS postulate that $\Delta EJF\cong \Delta GJQ$ .

Corresponding parts of congruent triangles are congruent so $\overline{EF}\cong \overline{GQ}$ .

Given that $\overline{EH}\cong \overline{PH}$ , also given that $\overline{EH}$ is a median and a median from $E$ to the opposite $\overline{FG}$ bisects $\overline{FG}$ hence, $\overline{FH}\cong \overline{GH}$ . Vertically opposite angles are equal so $\angle EHF\cong \angle PHG$ . From these properties it can be concluded by SAS postulate that $\Delta EHF\cong \Delta PHG$ .

Corresponding parts of congruent triangles are congruent so $\overline{EF}\cong \overline{PG}$ .

From $\overline{EF}\cong \overline{GQ}$ and $\overline{EF}\cong \overline{PG}$ , using the transitive property, $\overline{GQ}\cong \overline{PG}$ .

b.

To determine

To prove :  That $\overline{GQ}$ and $\overline{GP}$ are parallel to $\overline{EF}$ .

Explanation of Solution

Given information :  Consider a scalene $\Delta EFG$ . $\overline{EH}$ is a median and $P$ is a point on $\overrightarrow{EH}$ such that $\overline{EH}\cong \overline{HP}$ . Also, $\overline{FJ}$ is a median and $Q$ is a point on $\overrightarrow{FJ}$ such that $\overline{FJ}\cong \overline{JQ}$ . It is to be proved in paragraph form that $\overline{GQ}\parallel \overline{EF}$ and $\overline{GP}\parallel \overline{EF}$ .

  

Formula used :  If a pair of alternate interior angles iscongruent then two lines are parallel.

Proof :  From the SAS postulate it was concluded that $\Delta EHF\cong \Delta PHG$ .

Since corresponding parts of congruent triangles are congruent, $\angle HEF=\angle HPG$ . Now $\angle HEF$ are pair of alternate interior angles and $\overrightarrow{EP}$ is a transversal so $\overline{GP}\parallel \overline{EF}$ . Also $\overline{GQ}\cong \overline{GP}$ hence it can also be said that $\overline{GQ}\parallel \overline{EF}$ .

Hence $\overline{GP}\parallel \overline{EF}$ and $\overline{GQ}\parallel \overline{EF}$ .

c.

To determine

To prove :  The points $P$ , $Q$ , and $G$ are collinear.

Explanation of Solution

Given information :  Consider a scalene $\Delta EFG$ . $\overline{EH}$ is a median and $P$ is a point on $\overrightarrow{EH}$ such that $\overline{EH}\cong \overline{HP}$ . Also, $\overline{FJ}$ is a median and $Q$ is a point on $\overrightarrow{FJ}$ such that $\overline{FJ}\cong \overline{JQ}$ . It is to be proved in paragraph form that points $P$ , $Q$ , and $G$ are collinear.

Formula used :  Three points $A$ , $B$ ,and $C$ are collinear if they lie on the same line and $\overline{AB}+\overline{BC}\cong \overline{AC}$ .

If two lines are parallel to the same line then they are parallel to each other.

Proof :  Since $\Delta EHF\cong \Delta PHG$ and $\Delta EJF\cong \Delta GJQ$ it was proved using transitive property that $\overline{GQ}\cong \overline{GP}$ . Consider the points $P$ , $Q$ and $G$ ,it was shown that $\overline{GP}\parallel \overline{EF}$ and $\overline{GQ}\parallel \overline{EF}$ .

Now,two lines parallel to the same line are also parallel to each other $\overline{GQ}\parallel \overline{GP}$ .

Thus two lines $\overline{GQ}$ and $\overline{GP}$ are parallel to each other with the same magnitude so the points $P$ , $Q$ and $G$ lie on the same line and hence are collinear.