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Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.
Since Δ ADF ≅ Δ DFC
∠ADF = ∠CDF
∴ ∠ADC = 2 ∠CDF
Similarly we can prove ∠CEB = 2∠CEF
Since || m
∠ADC + ∠CEB = 180o
⇒2∠CDF + 2∠CEF = 180o
⇒ ∠CDF + ∠CEF = 90o
In Δ DFE
∠DFE = 90o
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3x+3/x2 -6x+5
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fig angles of a irregular polygons exterior and interior .
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