State true or false.
In the figure given below, PA and PB are tangents to the circle from point P, Where PA = PB.

True

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False

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Solution

The correct option is A True

Join the points OP,OA and OB
Now, in triangles OPA and OPB,
OP = OP (common side)
OA = OB (radii)
$∠ O A P = ∠ O B P = 90^{\circ}$ (Tangent is perpendicular to the radius)
Hence, $Δ O A P ≅ Δ O B P$ (RHS congruency)
Hence, PA = PB (CPCT)

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?
State true or false.

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State true or false. In the figure given below, PA and PB are tangents to the circle from point P, Where PA = PB.

The correct option is A True Join the points OP,OA and OB Now, in triangles OPA and OPB, OP = OP (common side) OA = OB (radii) ∠OAP=∠OBP=90∘ (Tangent is perpendicular to the radius) Hence, ΔOAP≅ΔOBP (RHS congruency) Hence, PA = PB (CPCT)

State true or false.
In the figure given below, PA and PB are tangents to the circle from point P, Where PA = PB.

The correct option is A True

Join the points OP,OA and OB
Now, in triangles OPA and OPB,
OP = OP (common side)
OA = OB (radii)
$∠ O A P = ∠ O B P = 90^{\circ}$ (Tangent is perpendicular to the radius)
Hence, $Δ O A P ≅ Δ O B P$ (RHS congruency)
Hence, PA = PB (CPCT)