In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show that

(i) BD > AD

(ii) DC > AD

(iii) AC > DC

(iv) AB > BD.

In △ABC,

∠B = 30° and ∠C = 40°

∠BAC = 180° - (30° + 40°) = 180° - 70° = 110°.

As, AD is bisector of ∠A,

∴ ∠BAD = ∠CAD = $\dfrac{110°}{2} = 55°$.

(i) Now in △ABD,

∠BAD > ∠B

∴ BD > AD.

Hence, proved that BD > AD.

(ii) Now in △ACD,

∠CAD > ∠C

∴ DC > AD.

Hence, proved that DC > AD.

(iii) Now in △ACD,

∠ADC = 180° - (40° + 55°) = 180° - 95° = 85°.

In △ACD,

∠ADC > ∠CAD

∴ AC > DC.

Hence, proved that AC > DC.

(iv) In △ADB,

∠ADB = 180° - ∠ADC = 180° - 85° = 95°.

Thus, ∠ADB > ∠BAD

∴ AB > BD.

Hence, proved that AB > BD.