In a circle, chords PQ and TR are parallel. Q is joined to a point S on the circumference which lies between T and R such that <PQT = <TQS. Prove
that TR = QS.​

Step-by-step explanation:

As OQ⊥ PR

(radius ⊥ tangent) & AB∣∣PR

⇒AB⊥OL

and perpendicular from centre to chord bisects the chord.

⇒AL=BL,

∠QLB=∠QLA

& LQ=LQ\Rightarro$$ By SAS

congruency, ΔQLA≅ΔQLB

⇒∠AQL=∠BQL=20

o

(90

o

−70

o

)

⇒∠AQB=40

o

.

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.