Chord and Tangent Intersect Externally
Here we will prove that if a chord and a tangent intersect externally then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.
Given: XY is a chord of a circle and TZ is a tangent that touches the circle at T. XY is produced to cut the tangent TZ at Z.
To prove: XZ × YZ = TZ2.
Construction: Join T to X and Y.
Proof:
|
Statement |
Reason |
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1. In ∆XTZ and ∆YTZ, (i) ∠XZT = ∠YZT (ii) ∠TXZ = ∠YTZ |
1. (i) Common angle. (ii) Angle between chord and tangent is equal to the angle in the alternate segment. |
|
2. In ∆XTZ ∼ ∆YTZ |
2. By AA criterion for similarity. |
|
3. \(\frac{XZ}{TZ}\) = \(\frac{TZ}{YZ}\) ⟹ XZ × YZ = TZ^2 (proved). |
3. Corresponding sides of similar triangles are proportional. |
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