D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.

In △ ADC,

⇒ AD > AC (Given)

∴ ∠ACD > ∠ADC (If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.) …….(1)

In △ ABD,

⇒ ∠ADC = ∠ABD + ∠BAD (An exterior angle is equal to sum of two opposite interior angles) ………(2)

Substituting value of ∠ADC from equation (2) in (1), we get :

⇒ ∠ACD > ∠ABD + ∠BAD ……(3)

∴ ∠ACD > ∠ABD …..(4)

From figure,

⇒ ∠ACD = ∠ACB and ∠ABD = ∠ABC

Substituting above values in equation (4), we get :

⇒ ∠ACB > ∠ABC

Thus, in △ ABC,

⇒ AB > AC (If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.)

Hence, proved that AB > AC.