prove that the angle between the two tangents draw from an external point to a circle is supplementary to the angle substended by the line segment joining the point of contact at the centre Related: Tangent to a Circle - Circles, CBSE, Class 10, Mathematics

Question

prove that the angle between the two tangents draw from an external point to a circle is supplementary to the angle substended by the line segment joining the point of contact at the centre Related:  Tangent to a Circle - Circles, CBSE, Class 10, Mathematics

Answer

Answer

Hiii

To prove: BPA+BOA=180

PROOF -: quadrilateral AOBP

PBO=90∘ (as the angle between the radius and the tangent to the circle is 90∘).

Similarly, PAO=90∘

PAO+PBO+BOA+BPA=360∘ (the sum of the angles of a quadrilateral)

Therefore 90∘+90∘+BOA+BPA=360

BOA+BPA=360∘−180∘=180

HENCE PROVED....

Answer

￼

Answer

Statement: The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the point of contact at the center.

Proof:

Given:
- Let O be the center of the circle.
- P be the external point from which two tangents PT1 and PT2 are drawn to the circle.
- A and B be the points of contact of the tangents PT1 and PT2 with the circle.
- ΔOAB is the triangle formed by the center of the circle and the points of contact of the tangents.

To prove:
∠PT1O + ∠PT2O = 180°

Proof:

Step 1: ΔOAB is an isosceles triangle.
- The tangents drawn from an external point to a circle are equal in length.
- Therefore, PA = PB.

Step 2: OP is the perpendicular bisector of AB.
- The line joining the center of a circle to the midpoint of a chord is perpendicular to the chord.
- Therefore, OP is perpendicular to AB and bisects it at point M.

Step 3: ΔPOM is congruent to ΔPOB.
- OM is common to both triangles.
- OP = OP (common side).
- ∠POM = ∠POB = 90° (both are right angles).

Step 4: ∠PT1O = ∠PT2O.
- In ΔPOM, ∠POM = ∠POM (common angle).
- ∠PT1O = ∠PT2O (corresponding angles in congruent triangles).

Step 5: ∠PT1O + ∠PT2O = 180°.
- ∠PT1O = ∠PT2O (proved in step 4).
- Therefore, the sum of the angles is 2∠PT1O.
- 2∠PT1O = 180° (angles around a point sum up to 360°).
- ∠PT1O = 180°/2 = 90°.

Conclusion:
The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the point of contact at the center.
∠PT1O + ∠PT2O = 180°.

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