Draw BE⊥PQ and CF⊥RS.
∠1 = ∠2 …(i)[∵ Angle of incident = Angle of reflection]
Similarly,
∠3 = ∠4 …(ii)
and,
∠2 = ∠3 …(iii)[∵ Alternate Angles]
⇒ ∠1 = ∠4 [From the equations (i), (ii) and (iii)]
⇒ 2∠1 = 2∠4
⇒ ∠1 + ∠1 = ∠4 + ∠4
⇒ ∠1 + ∠2 = ∠3 + ∠4 [From the equation (i) and (ii)]
⇒ ∠BCD = ∠ABC
Since, the alternate angles are equal. Hence, AB||CD.

∠PRS is the exterior angle of △PQR.
Therefore,
∠PRS = ∠QPR + ∠PQR
⇒ 1/2 ∠PRS = 1/2 ∠QPR + 1/2 ∠PQR
⇒ ∠TRS = 1/2 ∠QPR + ∠TQR …(1)[∵ ∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR]
∠TRS is exterior angle of △TQR
Therefore,
∠TRS = ∠QTR + ∠TQR …(2)
From, the equations (1) and (2), we have
∠QTR + ∠TQR = 1/2 ∠QPR + ∠TQR
⇒ ∠QTR = 1/2 ∠QPR

Given: y:z = 3:7
Let, y = 3k, therefore z = 7k
y = ∠1 = 3k [∵ Vertically Opposite Angles]
Given: CD||EF,
Therefore,
∠1 + z = 180° [∵ Sum of co-interior angles]
⇒ 3k + 7k = 180°
⇒ 10k = 180°
⇒ k = 180°/10 = 18°
Hence, z = 7k = 7 × 18° = 126°
Given that : AB||CD and CD||EF, therefore AB||EF
⇒ x = z = 126° [∵ Alternate Angles]

Contruction: Produce PQ, so that it intersect ST at M.
Given that: PQ||ST, therefore
∠1 = ∠2 [∵ Correspomding Angles]
⇒ ∠2 = 130°
∠2 + ∠3 = 180° [∵ Linear pair]
⇒ 130° + ∠3 = 180°
⇒ ∠3 = 180° – 130° = 50°
∠5 + ∠4 = 180° [∵ Linear pair]
⇒ 110° + ∠4 = 180°
⇒ ∠4 = 180° – 110° = 70°
In triangle QMR,
∠3 + ∠4 + ∠R = 180°
⇒ 50° + 70° + ∠R = 180°
⇒ 120° + ∠R = 180°
⇒ ∠R = 180° – 120° = 60°