Chapter 5.3, Problem 20WE

To determine

To prove that $R$ is equidistant from $U$ and $Q$ .

Explanation of Solution

Given:

  

  $\overrightarrow{PQ},\overrightarrow{RS},$ and $\overrightarrow{TU}$ are each perpendicular to $\overleftrightarrow{UQ}$ .

  $R$ is the midpoint of $\overrightarrow{PT}$ .

  $\Rightarrow TR=RP$

Concept used:

Corresponding angles of parallel line are equal.

Congruency of triangles.

Proof:

Join $RU$ and $RQ$ .

  

Since, $\overrightarrow{PQ},\overrightarrow{RS},$ and $\overrightarrow{TU}$ are each perpendicular to $\overleftrightarrow{UQ}$ .

So, $\angle TUS={90}^0$ and $\angle RSQ={90}^0$

  $\Rightarrow TU$ is parallel to $RS$ [Corresponding angles are equal ]

Similarly, $\angle RSQ={90}^0$ and $\angle PQV={90}^0$

  $\Rightarrow RS$ is parallel to $PQ$ [Corresponding angles are equal]

Therefore, $TU$ is parallel to $RS$ is parallel to $PQ$ .

Now, since $TU$ is parallel to $RS$ ,

  $\Rightarrow TR=US$ ….(i).

Now, since $RS$ is parallel to $PQ$ ,

  $\Rightarrow RP=SQ$ ….(ii).

Now, $TR=RP$ [given] …..(iii).

From (i), (ii), and (iii),

  $TR=US=RP=SQ$

  $\Rightarrow US=SQ$

Now in $\Delta URS$ and $\Delta QRS,$

  $US=SQ$ [Proved above]

  $\angle RSU=\angle RSQ={90}^0$ [Given]

  $RS=RS$ [Common]

Therefore, $\Delta URS\cong \Delta QRS$ by Side-Angle-Side congruency criteria.

  $\Rightarrow RU=RQ$ [ $\because$ Corresponding part of congruent triangles are congruent(equal)]

Therefore, R is equidistance from $U$ and $Q$ .

Hence Proved.

Conclusion:

Therefore, R is equidistance from $U$ and $Q$ .