9.2 CHAPTER 9. GEOMETRY
- AP = PB (given)
- OA = OB (radii)
- OP is common to both triangles.
∴�OAP≡�OBP (SSS).
OPAˆ = OPBˆ
OPAˆ +OPBˆ = 180◦ (APB is a straight line)
∴OPAˆ = OPBˆ = 90◦
∴ OP⊥ ABTheorem 3. The perpendicular bisector of a chord passes through the centre of thecircle.
Proof:
�A B
Q
P
Consider a circle. Drawa chord AB. Draw a line PQ perpendicular to AB such that PQ bisects AB
at point P. Draw lines AQ and BQ.
The aim is to prove that Q is the centre of the circle, by showing that AQ = BQ.
In�OAP and�OBP ,
- AP = PB (given)
2.∠QPA =∠QPB (QP⊥ AB)
- QP is common to both triangles.
∴�QAP≡�QBP (SAS).
From this, QA = QB. Since the centre of a circle is the only point inside a circle that has points on
the circumference at anequal distance from it, Q must be the centre of the circle.