Question

In the following figure, OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q.
Prove that:
( i ) ΔOPA ≅ ΔOQC 
( ii ) ΔBPC ≅ ΔBQA

Answer 1

(i) In ΔOPA and ΔOQC,
 OP = OQ                          ....[ radii of same circle ]
∠AOP = ∠COQ                  ... [ both 90° ] 
OA = OC                           ... [ sides of the square ]

By Side- Angle - Side criterion of congruence.
∴ ΔOPA ≅ ΔOQC             ...[ by SAS ]

(ii) Now, OP = OQ           ...[ radii ]
 and  OC = OA                 ...[ sides of the square ]
∴ OC - OP = OA - OQ  
⇒ CP = AQ                      ....(i)

In ΔBPC and ΔBQA,
BC = BA                           ...[ sides of the square ]
∠PCB = ∠QAB                 ...[ both 90° ]
 PC = QA                         ...[ by ( i ) ]

By Side- Angle-Side criterion of congruence,
∴ ΔBPC ≅ ΔBQA                ...[ by SAS ]