91Ó°ÊÓ

Here $\stackrel{→}{PQ}$, $\stackrel{→}{RS}$and $\stackrel{→}{TU}$are each perpendicular to $\stackrel{↔}{UQ}$.

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

Corresponding angles of a parallel line are equal.

Congruency of triangles.

Join RU and RQ:

Since, $\stackrel{→}{PQ}$, $\stackrel{→}{RS}$and $\stackrel{→}{TU}$are each perpendicular to $\stackrel{↔}{UQ}$.

So, $∠TUS={90}^{∘}$and $∠RSQ={90}^{∘}$.

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

Similarly, $∠RSQ={90}^{∘}$and $∠PQV={90}^{∘}$.

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 $\mathrm{Δ}URS$and $\mathrm{Δ}QRS$,

$US=SQ$[Proved above]

$∠RSU=∠RSQ={90}^{∘}$ [Given]

$RS=RS$ [Common]

Therefore, $\mathrm{Δ}URSâ‰\dots \mathrm{Δ}QRS$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).