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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
x^2-5x+4 can written in roots as (x-1)*(x-4) x^2-4 can be written interms of (x-2)(x+2).so [(x-1)(x-4)/(x-2)(x+2)]
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IS SQUARE A UNIQUE RHOMBUS?
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