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

Answer:

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′ ⇒∠AOM=∠BOM......(i)

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we have

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)⇒ΔAOM≅ΔBOM (SAS congruncy)

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)⇒ΔAOM≅ΔBOM (SAS congruncy)⇒AM=BM and∠AMO=∠BMO

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)⇒ΔAOM≅ΔBOM (SAS congruncy)⇒AM=BM and∠AMO=∠BMOBut

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)⇒ΔAOM≅ΔBOM (SAS congruncy)⇒AM=BM and∠AMO=∠BMOBut∠AMO+∠BMO=180°

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)⇒ΔAOM≅ΔBOM (SAS congruncy)⇒AM=BM and∠AMO=∠BMOBut∠AMO+∠BMO=180°⇒2∠AMO=180°

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)⇒ΔAOM≅ΔBOM (SAS congruncy)⇒AM=BM and∠AMO=∠BMOBut∠AMO+∠BMO=180°⇒2∠AMO=180°⇒∠AMO=90°

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)⇒ΔAOM≅ΔBOM (SAS congruncy)⇒AM=BM and∠AMO=∠BMOBut∠AMO+∠BMO=180°⇒2∠AMO=180°⇒∠AMO=90°Thus,AM=BM and∠AMO=∠BMO=90°

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)⇒ΔAOM≅ΔBOM (SAS congruncy)⇒AM=BM and∠AMO=∠BMOBut∠AMO+∠BMO=180°⇒2∠AMO=180°⇒∠AMO=90°Thus,AM=BM and∠AMO=∠BMO=90°HenceOO

′ ⇒∠AOM=∠BOM......(i)Now in ΔAOM and ΔBOM we haveOA=OB (radii of same circle)∠AOM=∠BOM (from (i))OM=OM (common side)⇒ΔAOM≅ΔBOM (SAS congruncy)⇒AM=BM and∠AMO=∠BMOBut∠AMO+∠BMO=180°⇒2∠AMO=180°⇒∠AMO=90°Thus,AM=BM and∠AMO=∠BMO=90°HenceOO ′

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Answer 2

Answer:

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