Question

PQ is a tangent to a circle with Centre O at the point Q a chord QA is drawn parallel to po. If AOB is a diameter of the circle prove that PB is the tangent to a circle at the point B​

Answer 1

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Answer 2

PB is the tangent to the circle at point B

Step-by-step explanation:

given : AQ ║ OP

proof : since AQ ║ OP

$\angle POQ = \angle OQA$ (alternate  interior)

$\angle OAQ = \angle POB$ (corresponding angle )

OA = OQ  (radii)

$\angle OQA = \angle OAQ$  (angle opp to equal sides)

therefore ,

$\angle POQ = \angle POB$

now , in triangle OPB and OPQ

$\angle POQ = \angle POB$ (proved above )

OB = OQ (radii)

OP= PO (common)

thus

$\bigtriangleup OPQ \cong \bigtriangleup OPB$

therefore

$\angle OBP = \angle OQP$

(by CPCT)

and  $\angle OBP = 90^\circ$

and OB ⊥ BP

since , Radius is perpendicular to the tangent at point B

therefore ,

PB is the tangent to the circle at point B

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