∠QRT = ∠RQS + ∠QSR
| ∵ The exterior angle is equal to the sum of the two interior opposite angles. ⇒ 65°
= 28° + ∠QSR
⇒ ∠QSR = 65° - 28° = 37°
∵ PQ ≠ SP
∴ ∠QPS = 90°
∵ PQ || SR
∴ ∠QPS + ∠PSR = 180°
| ∵ The sum of consecutive interior angles on the same side of the transversal is 180°
⇒ 90° + ∠PSR = 180°
⇒ ∠PSR = 180° - 90° = 90°
⇒ ∠PSQ + ∠QSR = 90°
⇒ y + 37° = 90°
⇒ y = 90° - 37° = 53°
In ∆PQS,
∠PQS + ∠QSP + ∠QPS = 180°
| ∵ The sum of all the angles of a triangle is 180°
⇒ x + y + 90° = 180°
⇒ x + 53° + 90° = 180°
⇒ x + 143° = 180°
⇒ x = 180° - 143° = 37°.
The side EF, FD and DE of a triangle DEF are produced in order forming three exterior angles DFP, EDQ and FER respectively. Prove that
∠DFP + ∠EDQ + ∠FER = 360°.
OR
Prove that the sum of the exterior angles of a triangl is 360°.