Question

The given figure shows a circle with centre
0. P is mid-point of chord AB.
Show that OP is perpendicular to AB.​

Answer 1

Construction(s): Join OA and OB (See: Figure)

Obtained Figure: Triangle OAP and OBP

Now, in ∆OAP and OBP:

OA = OB (Radius of circle)

AP = BP (P is the mid point of chord and bisects it into two equal parts)

OP = OP (Common)

Thus ∆OAP is congruent to ∆OBP.

Automatically,

∠OPA = ∠OPB = 90° (Proof Below)

Proof:

∠OPA+∠OPB=180 (Linear Pair)

but,

∠OPA=∠OPB (CPCT)

Thus 2∠OPA or 2∠OPB= 180

∠OPB=∠OPA=180/2

Or ∠OPB=∠OPA=90°

Hence proved.

Answer 2

Answer:

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