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

.

idk

step-by-step explanation:

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option