Question

6. The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Prove it.

Answer 1

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Answer 2

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$\sf\underbrace{Question: }$

• The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.??

$\sf\underline\red{Given }$

• A Circle with centre at O.
• AB is chord of circle & OX bisects AB i,e. AX = BX

$\sf\underline\orange{To\:Prove:}$

• $\sf{OX \:⊥\: AB}$

$\sf\underline\purple{Solution:}$

In ∆AOX & ∆BOX

• $⟹$$\sf{OA \:= \:OB}$

• $⟹$$\sf{OX\:=\:OX}$

• $⟹$$\sf{AX\:=\:BX}$

• $⟹$$\sf{∴△AOX\: ≈ \:△BOX}$

• $⟹$$\sf{∠AXO\: ≈\: BXO\:(CPCT)...1}$

In line AB

Hence, $\sf{∠AXO\:∠BXO\: From\: Linear\:Pair}$

• $⟹$$\sf{∠AXO\:+\:∠BXO\:=\:180°}$

• $⟹$$\sf{∠AXO\:+\:∠AXO\:=\:180°}$

• $⟹$$\sf{2∠AXO\:=\:180°}$

• $⟹$$\sf{∠AXO\:=}$$\sf\dfrac{180°}{2}$

• $⟹$$\sf{∠AXO\:=\:90°}$

• $⟹$$\sf{∴∠AXO\:=\:∠BXO\:90°}$

• $⟹$$\sf{=\:OX \:⊥\: AB}$

$\bf\underline{Hence,\:proved}$

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