Prove that if the arms of an angle are respectively perpendicular to the arms of another angle, then the angles are either equal or supplementary.
(i) If the arms of an angle are respectively perpendicular to the arms of another angle, then the angles will be equal:
(ii) If the arms of an angle are respectively perpendicular to the arms of another angle, then the angles will be supplementary:

(i) 

Given: AB and BC are the arms of an ∠ ABC. EQ and QD are the arms of another angle.
∠ QEO = 90° , ∠ ODB = 90° .
To Prove: ∠ OBD = ∠ EQO
Proof:
In Δ BOD, ∠ OBD + ∠ BOD + ∠ ODB = 180° [Sum of the three angles of a triangle is 180° ]
⇒ ∠ OBD + ∠ BOD + 90° = 180°
⇒ ∠ OBD + ∠ BOD = 90° …………………(i)
In Δ EQO, ∠ EQO + ∠ QOE + ∠ OEQ = 180° [Sum of the three angles of a triangle is 180° ]
⇒ ∠ EQO + ∠ QOE + 90° = 180°
⇒ ∠ EQO + ∠ QOE = 90° …………………(ii)
From (i) and (ii),
∠ OBD + ∠ BOD = ∠ EQO + ∠ QOE
But, ∠ BOD = ∠ QOE [Vertically opposite angles]
⇒ ∠ OBD = ∠ EQO
(ii)

Given:
∠ ABC and ∠ PQR are two angles whose arms are respectively perpendicular to each other respectively.
AB ⊥ QR and BC ⊥ QP.
To Prove: ∠ DBE and ∠ EQD are supplementary.
Construction: Join BQ.
Proof:
In Δ BDQ, ∠ DBQ + ∠ BQD + ∠ QDB = 180° [Sum of the three angles of a triangle is 180° ]
⇒ ∠ DBQ + ∠ BQD + 90° = 180°
⇒ ∠ DBQ + ∠ BQD = 90° ………………………..(i)
In Δ BQE, ∠ EBQ + ∠ BQE + ∠ BEQ = 180° [Sum of the three angles of a triangle is 180° ]
∠ EBQ + ∠ BQE + 90° = 180°
∠ EBQ + ∠ BQE = 90° ………………………….(ii)
Adding (i) and (ii),
⇒ (∠ DBQ + ∠ BQD) + (∠ EBQ + ∠ BQE) = 90° + 90°= 180°                                                      
⇒ ∠ DBE + ∠ EQD = 180°
∴ ∠ DBE and ∠ EQD are supplementary.