Exercise 14 Page 149

We will construct a segment that is twice as long as $\overline{PQ}.$

We will first draw a segment $\overline{QR}$ that is congruent to $\overline{PQ}.$ To do that we will first draw a ray extending $\overline{PQ}$ to the right.

Next, we will measure $PQ$ using our compass.

With the same compass setting, we will put the compass point on point $Q$ and draw an arc that intersects the ray.

Finally, we will label the point of intersection as $R$ and remove the unnecessary part.

Now, we will prove that $PR=2PQ$ given that $\overline{PQ}\cong \overline{QR}.$ As always, we will start with stating the given information and the statement to be proven. In order to have an idea about the lengths of the given segments, we will use the Definition of Congruent Segments. The definition states that two segments are congruent if and only if they have the same length. When we look at the figure, we see that the points $P,$ $Q,$ and $R$ are collinear and $Q$ is between $P$ and $Q$. In this case, we can use the Segment Addition Postulate to write our next step. Since we know that $PQ$ is equal to $QR,$ we will substitute $PQ$ for $QR$ into the given equation. Finally, we will add the terms and complete the proof. Combining these steps, let's construct a two-column proof.

Statements	Reasons
$\overline{PQ}\cong \overline{QR}$	Given
$PQ=QR$	Definition of Congruent Segments
$PR=PQ+QR$	Segment Addition Postulate
$PR=PQ+PQ$	Substitution Property of Equality
$PR=2PQ$	Addition Property of Equality