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°

Answer 1

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°.