In the following figure, ∠BAC = 60° and ∠ABC = 65°. Prove that :

(i) CF > AF

(ii) DC > DF

(i) In △ BEC,

⇒ ∠CBE + ∠BEC + ∠BCE = 180°

⇒ 65° + 90° + ∠BCE = 180°

⇒ ∠BCE = 180° - 90° - 65° = 25°.

From figure,

⇒ ∠DCF = ∠BCE = 25° …….(1)

In △ CDF,

⇒ ∠DCF + ∠FDC + ∠CFD = 180°

⇒ 25° + 90° + ∠CFD = 180°

⇒ ∠CFD = 180° - 90° - 25° = 65° ………(2)

From figure,

AFD is a straight line,

⇒ ∠AFC + ∠CFD = 180°

⇒ ∠AFC + 65° = 180°

⇒ ∠AFC = 180° - 65° = 115° ……..(3)

In △ ACE,

By angle sum property of triangle,

⇒ ∠ACE + ∠CEA + ∠EAC = 180°

⇒ ∠ACE + ∠CEA + ∠BAC = 180° (From figure, ∠EAC = ∠BAC)

⇒ ∠ACE + 90° + 60° = 180°

⇒ ∠ACE + 150° = 180°

⇒ ∠ACE = 180° - 150° = 30° ……..(4)

In △ AFC,

By angle sum property of triangle,

⇒ ∠AFC + ∠ACF + ∠FAC = 180°

⇒ 115° + ∠ACE + ∠FAC = 180° (From figure, ∠ACF = ∠ACE)

⇒ 115° + 30° + ∠FAC = 180°

⇒ ∠FAC + 145° = 180°

⇒ ∠FAC = 180° - 145° = 35° ……..(5)

In △ AFC,

⇒ ∠FAC > ∠ACF

∴ CF > AF (If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.)

Hence, proved that CF > AF.

(ii) In △ CDF,

⇒ ∠DCF = 25° ……..[From equation (1)]

⇒ ∠CFD = 65° ……..[From equation (2)]

⇒ ∠CFD > ∠DCF

∴ DC > DF (If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.)

Hence, proved that DC > DF.