In the figure (1) given below, AD = BD = DC and ∠ACD = 35°. Show that

(i) AC > DC

(ii) AB > AD.

From figure,

∠DAC = ∠ACD = 35° (As angles opposite to equal sides are equal.)

∠ADC = 180° - (35° + 35°) = 180° - 70° = 110°.

Since, ∠ADC > ∠DAC

∴ AC > DC (Side opposite to greater angle is greater.)

Hence, proved that AC > DC.

(ii) From figure,

∠ADB = 180° - ∠ADC = 180° - 110° = 70°.

Considering △ABD,

AD = BD.

∴ ∠BAD = ∠ABD = a.

⇒ a + a + ∠ADB = 180°

⇒ 2a + 70° = 180°

⇒ 2a = 110°

⇒ a = 55°.

Since, ∠ADB > ∠ABD

∴ AB > AD. (Side opposite to greater angle is greater.)

Hence, proved that AB > AD.