Question

Prove that if a line segment from the centre of a circle to a chord bisects the chord is perpendicular on chord,​

Answer 1

Correct Question :-

• Prove that the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.

Given :-

• O is the center of circle
• AB is chord of circle
• OX bisects AB   i.e.   AX = BX

To Prove :-

• OX ⊥ AB

Explanation :-

➠ In ∆AOX and ∆BOX,

$\implies \: {\mathrm{OA =OB\: \: \:(Radius\:of\:Circle)}$

$\implies \: {\mathrm{OX =XO\: \: \:(Common)}$

$\implies \: {\mathrm{AX =BX\: \: \:(Given)}$

$\therefore \: {\mathrm{\bold{{\triangle AOX \cong \triangle BOX \: \: \: (SSS \;Rule)}}}$

$\implies \: {\boxed{\mathrm{\angle AXO= \angle BXO\: \: \:(C.P.C.T)\;\;\;\;....(1)}}$

➠ On line AB,

Hence, ∠AXO and ∠BXO form a linear pair

$\implies \: {\mathrm{{\angle AXO + \angle BXO = 180^{\circ}\: \: \:(L.P)}}$

$\implies \: {\mathrm{{\angle AXO + \angle AXO = 180^{\circ}\: \: \:(Equation\;1)}}$

$\implies \: {\mathrm{{2\angle AXO = 180^{\circ}}}$

$\implies \: {\mathrm{{\angle AXO = \dfrac{180^{\circ}}{2}}}$

$\implies \: {\boxed{\mathrm{{\bold{\angle AXO = 90^{\circ}}}}}$

$\therefore \; \;{{\boxed{\mathrm {\; \angle AXO = \angle BXO = 90^{\circ}}}}}$

$\implies \:{\underline{\boxed{\mathrm {\bold {OX \: \perp \: AB}}}}$

Hence, Proved.

@Agamsain