In the figure (1) given below, prove that (i) CF > AF (ii) DC > DF.

(i) In △ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 60° + 65° + ∠C = 180°

⇒ ∠C = 180° - 125° = 55°.

In △BEC,

⇒ ∠B + ∠E + ∠C = 180°

⇒ 65° + 90° + ∠C = 180°

⇒ ∠C = 180° - 155° = 25°.

In △DFC,

⇒ ∠D + ∠C + ∠F = 180°

⇒ 90° + 25° + ∠F = 180°

⇒ ∠F = 180° - 115° = 65°.

From figure,

∠AFE = ∠DFC = 65° (Vertically opposite angles are equal.)

An exterior angle is equal to the sum of two opposite interior angles.

⇒ ∠AFE = ∠FAC + ∠FCA

⇒ 65° = ∠FAC + 30°

⇒ ∠FAC = 65° - 30° = 35°.

In △AFC,

∠FAC = 35°

∠FCA = 30°

∴ ∠FAC > ∠FCA

∴ CF > AF (As side opposite to greater angle is greater.)

Hence, proved that CF > AF.

(ii) In △DFC,

∠DFC = 65°

∠DCF = 25°

∴ ∠DFC > ∠DCF

∴ DC > DF (As side opposite to greater angle is greater.)

Hence, proved that DC > DF.