From the following figure, prove that :

(i) ∠ACD = ∠CBE

(ii) AD = CE

(i) In Δ ACB,

⇒ AC = AB (Given)

∴ ∠ABC = ∠ACB (Angles opposite to equal sides are equal) ……..(1)

Since, DCB is a straight line.

∴ ∠ACD + ∠ACB = 180° ……(2)

Since, ABE is a straight line.

∴ ∠ABC + ∠CBE = 180° ……(3)

Equating equations (2) and (3), we get :

⇒ ∠ACD + ∠ACB = ∠ABC + ∠CBE

⇒ ∠ACD + ∠ACB = ∠ACB + ∠CBE [From equation (1)]

⇒ ∠ACD = ∠CBE.

Hence, proved that ∠ACD = ∠CBE.

(ii) In △ ACD and △ CBE,

⇒ DC = CB (Given)

⇒ AC = BE (Given)

⇒ ∠ACD = ∠CBE (Proved above)

∴ Δ ACD ≅ Δ CBE (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

⇒ AD = CE.

Hence, proved that AD = CE.