prove that tangent drawn at end of chord of a circle make equal angle with the chords
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Let PQ be the chord of a circle with center O
Let AP and AQ be the tangents at points P and Q respectively.
Let us assume that both the tangents meet at point A.
Join points O, P. Let OA meets PQ at R
Here we have to prove that ∠APR = ∠AQR
Consider, ΔAPR and ΔAQR
AP = AQ [Tangents drawn from an internal point to a circle are equal]
∠PAR = ∠QAR
AR = AR [Common side]
∴ ΔAPR ≅ ΔAQR [SAS congruence criterion]
Hence ∠APR = ∠AQR [CPCT]
Let AP and AQ be the tangents at points P and Q respectively.
Let us assume that both the tangents meet at point A.
Join points O, P. Let OA meets PQ at R
Here we have to prove that ∠APR = ∠AQR
Consider, ΔAPR and ΔAQR
AP = AQ [Tangents drawn from an internal point to a circle are equal]
∠PAR = ∠QAR
AR = AR [Common side]
∴ ΔAPR ≅ ΔAQR [SAS congruence criterion]
Hence ∠APR = ∠AQR [CPCT]
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