Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that angle PTQ = 2 times angle OPQ.
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Step-by-Step Solution
Step 1: Identify properties of tangents and radii
When a tangent is drawn to a circle, the radius drawn to the point of tangency is perpendicular to the tangent. Therefore, ∠OPT=90∘ and ∠OQT=90∘. Also, the lengths of tangents from an external point to a circle are equal, so TP=TQ.
Step 2: Analyze triangle TPQ
Since TP=TQ (tangents from an external point), triangle TPQ is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal. Let ∠PTQ=θ. Then, the sum of angles in △TPQ is 180∘, so ∠TPQ+∠TQP+∠PTQ=180∘. This implies 2∠TPQ+θ=180∘, so ∠TPQ=2180∘−θ=90∘−21θ.
Step 3: Relate angles using perpendicularity
We know that ∠OPT=90∘. From the diagram, we can see that ∠OPQ=∠OPT−∠TPQ. Substituting the values we found: ∠OPQ=90∘−(90∘−21θ)=90∘−90∘+21θ=21θ.
Step 4: Conclude the proof
Since we defined ∠PTQ=θ and found that ∠OPQ=21θ, we can conclude that ∠PTQ=2∠OPQ. This completes the proof.