Given: $\overline{\text{PQ}}$, $\overline{\text{RS}}$and $\overline{\text{TU}}$are each perpendicular to $\overleftrightarrow{\text{UQ}}$;

R is the midpoint of $\overline{\text{PT}}$;

Prove: R is equidistant from U and Q.

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

R is the midpoint of $\overrightarrow{PT}$. That implies $TR=RP$

Corresponding angles of a parallel line are equal.

Congruency of triangles.

Join RU and RQ:

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

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

That implies, TU is parallel to RS [ Corresponding angles are equal.]

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

That implies, 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,

That implies, $TR=US$ …… (i)

Now, since RS is parallel to PQ,

That implies, $RP=SQ$ …… (ii)

Now, $TR=RP$[given] …… (iii)

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

$TR=US=RP=SQ$

That implies, $US=SQ$.

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

$US=SQ$[Proved above]

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

$RS=RS$ [Common]

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

That implies $RU=RQ$[ Since corresponding part of congruent triangles are congruent (equal)].

Therefore, R is equidistant from U and Q.

Therefore, R is equidistant from U and Q (proved).