circle O and circle P intersect at A and B. Show that OP is the perpendicular bisector of the common chord AB

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.

95.03 is the answer. hope this !

i can answer this if i have the answer choices