Question

prove that the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord.​

Answer 1

Answer:

Since the line segment joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. Since AB and CD are equal chords, they are equidistant from the other. Hence Proved.

Answer 2

See the attachment ⬆️⬆️⬆️

$\huge \bigstar$$\huge\bold{\mathtt{\purple{A{\pink{N{\green{S{\blue{W{\red{E{\orange{R}}}}}}}}}}}}}$$\huge \Rightarrow$

$\large\bf \underline\red {Given}\Rightarrow$

seg AB is a chord of a circle with centre O and P is the midpoint of chord AB of the circle. That means seg AP $\cong$ seg PB.

$\large \bf \underline\red {To\:Prove}\Rightarrow$

seg OP $\perp$ chord AB.

$\large\bf \underline\red {Proof}\Rightarrow$

$\bf Draw \: seg \: OA \: and \: seg \: OB \\ \\ \bf In \: ∆AOP \: and∆BOP \\ \\\bf seg \: OA \: \cong seg OB \: \: \: \: \: ...(Raddi \: of \: the \: same \: circle \\ \bf seg \: OP \cong \: seg \: OP \: \: \: \: ...(Common \: side) \\ \bf seg \: AP \cong seg \: BP \: \: \: \: \: ...(Given) \\ \bf \therefore ∆AOP \cong \: ∆BOP \: \: \: \: ...(SSS \: test \: of \: congruence) \\ \bf \angle OPA\cong \angle OPB \: \: \: \: \: \: \: \: ... (c.a.c.t) \: \: ...(1) \\ \\\bf Now, \angle OPA +\angle OPB=180° \: \: \: ...(Angles \: in \: a \: linear \: pair) \\ \bf\therefore \angle OPB+\angle OPB=180° \: \: \: \: ...[From(1)] \\ \bf \therefore2\angle OPB=180° \\ \bf \therefore \angle OPB=90° \\ \bf\therefore seg \: OP \perp chord \: AB.$

Hence Proved