Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ=2∠OPQ. [Figure required: circle with centre O, external point T, tangents TP and TQ touching the circle at points P and Q.]
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Step-by-Step Solution
Step 1: Establish tangent properties
According to Theorem 10.1 in the NCERT textbook, the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, the radius OP is perpendicular to the tangent TP, and the radius OQ is perpendicular to the tangent TQ.
Step 2: Equal tangents from external point
According to Theorem 10.2 in the NCERT textbook, the lengths of tangents drawn from an external point to a circle are equal. Thus, the tangents TP and TQ from point T to the circle are equal in length.
Step 3: Identify isosceles triangle
Since TP=TQ, triangle TPQ is an isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are equal. Therefore, ∠TPQ=∠TQP.
Step 4: Apply angle sum property in △TPQ
The sum of angles in any triangle is 180∘. For triangle TPQ, the sum of its internal angles is 180∘. Substituting ∠TPQ=∠TQP, we get 2∠TPQ+∠PTQ=180∘.
Step 5: Express ∠PTQ in terms of ∠TPQ
From the angle sum property of △TPQ, we can express ∠PTQ as 180∘ minus two times ∠TPQ. This relation will be used later.
Step 6: Relate ∠TPQ to ∠OPQ
We know that OP⊥TP, so ∠OPT=90∘. This angle can be split into ∠OPQ and ∠TPQ. Hence, ∠OPQ+∠TPQ=90∘.
Step 7: Express ∠TPQ in terms of ∠OPQ
Rearranging the equation from the previous step, we can express ∠TPQ as 90∘−∠OPQ. This will allow us to substitute into our equation for ∠PTQ.
Step 8: Substitute and simplify
Substitute the expression for ∠TPQ from equation (2) into equation (1). This combines the relationships we've established to solve for ∠PTQ in terms of ∠OPQ.
Step 9: Final simplification
Distribute the 2 and simplify the expression. The 180∘ terms cancel out, leaving us with ∠PTQ=2∠OPQ, which proves the required statement.