it two circle intersect at two points prove that thay line on the perpendicular bisector of common chort
Answers
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
</div><div></div><div>But</div><div></div><div>
∠AMO+∠BMO=180
∘
⇒2∠AMO=180
∘
⇒∠AMO=90
∘
</div><div></div><div>Thus,AM=BMand\angle AMO = \angle BMO = {90^ \circ }</div><div></div><div>Hence,<span>OO'</span><span>isperpendicularbisectorofAB$$
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.
