Question

Hola guys!
Question--If two circles intersect in two points prove that the line through their centres is the perpendicular bisector of the common chord.
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Answer 1

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.

Answer 2

Answer:

The perpendicular bisected line of ab equal to 90 degree