Question

it two circle intersect at two points prove that thay line on the perpendicular bisector of common chort​

Answer 1

Step-by-step explanation:

Consider the problem

Consider two circle 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 centers.

Let OO

′

intersect AB at M

Now we draw line segments OA,OB,O'AandO'B</div><div><span></span></div><div>In\Delta OAO'\,and\,\Delta OBO',Wehave</div><div></div><div>

OA=OB(radiiofsamecircle)

O

′

A=O

′

B(radiiofsamecircle)

O

′

O=OO

′

(commonside)

</div><div>

⇒ΔOAO

′

≅ΔOBO

′

(SSScongruency)

⇒∠AOO

′

=∠BOO

′

⇒∠AOM=∠BOM......(i)

</div><div></div><div>NowIn\Delta AOM\,and\,\Delta BOM,Wehave</div><div></div><div>

OA=OB(radiiofsamecircle)

∠AOM=∠BOM(from(i))

OM=OM(commonside)

⇒ΔAOM≅ΔBOM(SAScongruncy)

⇒AM=BMand∠AMO=∠BMO

Answer 2

Given: Two circles, with centres O and O' intersect at two points A and B. A AB is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO' intersect AB at P.

To prove: OO' is the perpendicular bisector of AB.

Construction: Join OA, OB, O'A and O'B

Proof: In triangles OAO' and OBO', we have

OO' = OO' (Common)

OA = OB (Radii of the same circle)

O'A = O'B (Radii of the same circle)

⇒ △OAO' ≅ △OBO' (SSS congruence criterion)

⇒ ​∠AOO' = ∠BOO' (CPCT)

i.e., ∠AOP = ∠BOP

In triangles AOP and BOP, we have

OP = OP (Common)

∠AOP = ∠BOP (Proved above)

OA = OB (Radio of the same circle)

∴ △AOR ≅ △BOP (By SAS congruence criterion)

⇒ AP = CP (CPCT)

and ∠APO = ∠BPO (CPCT)

But ∠APO + ∠BPO = 180° (Linear pair)

∴ ∠APO + ∠APO = 180° ⇒ 2∠APO = 180°

⇒ ∠APO = 90°

Thus, AP = BP and ∠APO = ∠BPO = 90°

Hence, OO' is the perpendicular bisector of AB.