From the following figure; prove that :

(i) AB > BD

(ii) AC > CD

(iii) AB + AC > BC

(i) Since, BDC is a straight line.

∴ ∠ADB + ∠ADC = 180°

⇒ ∠ADB + 90° = 180°

⇒ ∠ADB = 180° - 90° = 90°.

In △ ABD,

∠BAD and ∠ABD will be definitely less than 90° as sum of angles of triangle equals to 180°.

∴ ∠ADB > ∠BAD

∴ AB > BD [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.] ………..(1)

Hence, proved that AB > BD.

(ii) From figure,

⇒ ∠ADC = 90°.

In △ ADC,

∠DAC and ∠DCA will be definitely less than 90° as sum of angles of triangle equals to 180°.

∴ ∠ADC > ∠DAC

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

Hence, proved that AC > CD.

(iii) Adding equations (1) and (2), we get :

⇒ AB + AC > BD + CD

⇒ AB + AC > BC.

Hence, proved that AB + AC > BC.