Chapter 6.4, Problem 1AE

a.

To determine

To prove: $l$ is the perpendicular bisector of $\overline{BC}$ .

Explanation of Solution

Given:

As from the given from figure the $\angle CPl$is $90°$ and $\angle BPl$is $90°$ the line $\overline{BC}$ is perpendicular to the line $l$ .

Also from the figure it is given that, $CP\cong BP$

Thus, line l is perpendicular bisector of $\overline{BC}$ .

Conclusion:

Therefore, $l$ is the perpendicular bisector of $\overline{AB}.$

b.

To determine

To prove: $SC=SB$ .

Explanation of Solution

Given:

Considering the triangle $\Delta SPC$ and $\Delta SPB,$

  $\Rightarrow \angle CPS=\angle BPS=90°$

  $\Rightarrow \overline{PC}=\overline{BP}$

  $\Rightarrow SP$ is the common side

Then, according to Side-Angle-Side theorem of congruency $\Delta SCP$ and $\Delta SBP$ are congruent.

So, now it can be said that $SC=SB$ .

Conclusion:

Therefore, $SC=SB$ is proved.

c.

To determine

To prove: $AS+SC=AC$ .

Explanation of Solution

Given:

From the given figure,

Considering the line segment $\overline{AC}$ , it is a straight line and segment $AC$ is made up of two segments, $AS\text{ and }SC$ and these two segments have a common point S .

So, $AS+SC=AC$

Conclusion:

As the segments AS and SC lie on a single straight line and having a common point, it can be said that $AS+SC=AC$ .

d.

To determine

To prove: $AS+SB=AC$ .

Explanation of Solution

Given:

From the part (b), it can be said that $\Delta SCP$ and $\Delta SBP$ are congruent.

So, $SC=SB$ and was proved that $AS+SC=AC$ in part (c).

  $\Rightarrow AS+SB=AC$

  $\Rightarrow AS+SB=AC$$\left[\because SC=SB\right]$

Conclusion:

Therefore, from the above statement it can be concluded that $AS+SB=AC$ .

e.

To determine

To prove: $XC=XB$ .

Explanation of Solution

Given:

Considering the triangle $\Delta XPC$ and $\Delta XPB,$

  $\Rightarrow \angle CPX=\angle BPX=90°$

  $\Rightarrow \overline{PC}=\overline{BP}$ [From part (b)]

  $\Rightarrow XP$ is the common side.

Then, according to Side-Angle-Side theorem of congruency $\Delta XCP$ and $\Delta XBP$ are congruent.

So, it can be said $XC=XB$ as corresponding sides of 2 congruent triangles are equal.

Conclusion:

Therefore $XC=XB$ .

f.

To determine

To prove: $AX+XC>AC$ .

Explanation of Solution

Given:

Considering the points A and C , there can be only 1 shortest line passing through these points.

Here line segment AC is the shortest straight line passing between these points.

Now point X lies outside the line AC , so if points A and C were connected by a line passing through any point outside the straight line be longer.

In simple words, line segment A-X-C will be longer than line segment A-S-C as the former segment passes through a point outside the shortest straight line between A and C .

Thus, $AX+XC>AC$ .

Conclusion:

As the point X lies outside the shortest straight line between points A and C segment $AX+XC>AC$ .

g.

To determine

To prove: $AX+XB>AS+SB$ .

Explanation of Solution

Given:

Calculation:

From the part (b), it was concluded that $SC=SB$ .

And from part (e), it was concluded that $XC=XB$ .

Now, as proven from part (f), $AX+XC>AC$ .

  $\begin{array}{l}AX+XB>AC\\ AX+XB>AS+SC\\ AX+XB>AS+SB\end{array}$ .

Conclusion:

By using the results from part (b), (e) and (f), it can be easily proved that

  $AX+XB>AS+SB$ .