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Question
In Fig. l and m are two parallel tangents at A and B. The tangent at C makes an intercept DE between n and m. Prove that ∠ DFE = 90°
Solution
In triangles ADF and DFC, we have DA = DC ...(Tangents drawn from external point are equal in length)
DF = DF ...(Common)
AF = CF ...(Radii of the same circle)
So, by SSS-Criterion of congruence, we get
Δ ADF = ΔDFC
⇒ ∠ADF = ∠CDF
⇒ ∠ADC = 2∠CDF ...(i)
Similarly, we can prove that
∠BEF = ∠CEF
⇒ ∠CEB = 2∠CEF ....(ii)
Now, ∠ADC + ∠CEB = 180° ...(Sum of the interior angles on the same side of transversal is 180°)
⇒ 2(∠CDF + ∠CDF) = 180°
⇒ ∠CDF + ∠CEF = `(180°)/2`
⇒ ∠CDF + ∠CEF = 90°
Now, in ∠DEF,
⇒ ∠DFE + ∠CDF + ∠CEF = 180°
⇒ ∠DFE + 90° = 180°
⇒ ∠DFE = 180° - 90°
⇒ ∠DFE = 90°.
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