In each of the following figures, write BC, AC and CD in ascending order of their lengths.

(i) In △ ABC,

⇒ AC = AB (Given)

⇒ ∠B = ∠C = 67°.

By angle sum property of triangle,

⇒ ∠BAC + ∠B + ∠C = 180°

⇒ ∠BAC + 67° + 67° = 180°

⇒ ∠BAC + 134° = 180°

⇒ ∠BAC = 180° - 134° = 46°.

In △ ABD,

By angle sum property of triangle,

⇒ ∠BAD + ∠B + ∠D = 180°

⇒ ∠BAD + 67° + 33° = 180°

⇒ ∠BAD + 100° = 180°

⇒ ∠BAD = 180° - 100° = 80°.

From figure,

⇒ ∠CAD = ∠BAD - ∠BAC = 80° - 46° = 34°.

We know that,

If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.

In △ ABC,

Since, ∠BAC < ∠ABC

⇒ BC < AC ……..(1)

In △ ACD,

Since, ∠CDA < ∠CAD

⇒ AC < CD ……..(2)

From equation (1) and (2), we get :

⇒ BC < AC < CD.

Hence, BC < AC < CD.

(ii) In △ ABC,

⇒ ∠BAC < ∠ABC

⇒ BC < AC [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.] …….(1)

By angle sum property of triangle,

⇒ ∠ACB + ∠BAC + ∠ABC = 180°

⇒ ∠ACB + 47° + 73° = 180°

⇒ ∠ACB + 120° = 180°

⇒ ∠ACB = 180° - 120° = 60°.

From figure,

As, BCD is a straight line.

⇒ ∠ACB + ∠ACD = 180°

⇒ 60° + ∠ACD = 180°

⇒ ∠ACD = 180° - 60° = 120°.

In △ ACD,

By angle sum property of triangle,

⇒ ∠ADC + ∠ACD + ∠CAD = 180°

⇒ ∠ADC + 120° + 31° = 180°

⇒ ∠ADC + 151° = 180°

⇒ ∠ADC = 180° - 151° = 29°.

Since, ∠ADC < ∠CAD, we get :

AC < CD [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.] …….(2)

From equations (1) and (2), we get :

⇒ BC < AC < CD.

Hence, BC < AC < CD.