Ray PQ and PRare perpendicular each other. Point B is the interior and Point A is the exterior of ∠QPR.
Ray PB and PA are perpendicular to each other.
(i) A pair of Complementary angles:
⇒ ∠QPB and ∠BPR because their sum is 90o
⇒ ∠BPR and ∠RPA because their sum is 90o
(ii) A pair of Supplementary angles:
⇒ ∠QPR and ∠BPA because their sum is 180o
(iii) A pair of congruent angles:
⇒ ∠BPA≡∠QPR [ Each angle is 90o ]
⇒ ∠BPA=∠QPR
So, ∠RPA+∠BPR=∠BPR+∠QPB [ Using angle addition property ]
So, ∠RPA=∠QPB
Hence, ∠RPA≡∠QPB