Answer:
Step-by-step explanation:
Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles and OO' is the line segment joining the centres
Let OO
′
intersect AB at M
Now Draw line segments OA,OB,O
′
AandO
′
B
In ΔOAO
′
andOBO
′
, we have
OA=OB (radii of same circle)
O
′
A=O
′
B (radii of same circle)
O
′
O=OO
′
(common side)
⇒ΔOAO
′
≅ΔOBO
′
(SSS congruency)
⇒∠AOO
′
=∠BOO
′
⇒∠AOM=∠BOM......(i)
Now in ΔAOM and ΔBOM we have
OA=OB (radii of same circle)
∠AOM=∠BOM (from (i))
OM=OM (common side)
⇒ΔAOM≅ΔBOM (SAS congruncy)
⇒AM=BM and∠AMO=∠BMO
But
∠AMO+∠BMO=180°
⇒2∠AMO=180°
⇒∠AMO=90°
Thus,AM=BM and∠AMO=∠BMO=90°
HenceOO is the perpendicular bisector of AB.
