The following figure shows a circle with centre O. If OP is perpendicular to AB,then AP $=$ BP.

True

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False

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Solution

The correct option is A True
Join $A O$ and $B O$
In $△ A O P, △ B O P$
$A O = B O$ $(∵ r a d i u s)$
$∠ O P A = ∠ O P B = 90^{\circ}$
$O P = O P$
By $S A S$ congruency
$⟹ △ A O P ≅ △ B O P$
$⟹ A P = B P$

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The following figure shows a circle with centre O. If OP is perpendicular to AB,then AP=BP.

The correct option is A TrueJoin AO and BOIn △AOP,△BOPAO=BO (∵radius)∠OPA=∠OPB=90∘OP=OPBy SAS congruency ⟹△AOP≅△BOP⟹AP=BP

The following figure shows a circle with centre O. If OP is perpendicular to AB,then AP $=$ BP.

The correct option is A True
Join $A O$ and $B O$
In $△ A O P, △ B O P$
$A O = B O$ $(∵ r a d i u s)$
$∠ O P A = ∠ O P B = 90^{\circ}$
$O P = O P$
By $S A S$ congruency
$⟹ △ A O P ≅ △ B O P$
$⟹ A P = B P$