In isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that :

(i) AC > AD

(ii) AE > AC

(iii) AE > AD

(i) In ∆ ABC,

⇒ AB = AC (Given)

⇒ ∠ACB = ∠ABC [Angles opposite to equal sides are equal] …….(1)

⇒ ∠ACD = ∠ABD [From figure, ∠ACB = ∠ACD and ∠ABC = ∠ABD] ………(2)

We know that exterior angle of a triangle is always greater than each of the interior opposite angle.

In ∆ ADC,

⇒ ∠ADB > ∠ACD

⇒ ∠ADB > ∠ABD [Using equation (2)]

⇒ AB > AD [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.]

⇒ AC > AD [As, AB = AC] …..(3)

Hence, proved that AC > AD.

(ii) We know that exterior angle of a triangle is always greater than each of the interior opposite angle.

In ∆ ACE,

⇒ ∠ACD > ∠AEC ……..(4)

From equation (2),

⇒ ∠ACD = ∠ABD

From figure,

⇒ ∠ABD = ∠ABE

⇒ ∠ACD = ∠ABE

⇒ ∠AEC = ∠AEB

Substituting value of ∠ACD and ∠AEC in equation (4), we get :

⇒ ∠ABE > ∠AEB

In △ AEB,

⇒ AE > AB [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.]

As, AB = AC,

∴ AE > AC

Hence, proved that AE > AC.

(iii) Since, AE > AC and AC > AD,

∴ AE > AD.

Hence, proved that AE > AD.