ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively such that AB = BE and AD = DF. Prove that △BEC ≅ △DCF.

Given,

AD = DF ….(1)

AB = BE ….(2)

We know that,

Opposite sides of parallelogram are equal.

∴ AD = BC ….(3)

∴ AB = CD ….(4)

From eq.(1) and (3), we get :

⇒ BC = DF

From eq.(2) and (4), we get :

⇒ BE = CD

We know that,

Opposite angles of a parallelogram are equal.

⇒ ∠ABC = ∠ADC = x (let)

From figure,

Since, AE is a straight line.

⇒ ∠CBE + ∠ABC = 180°

⇒ ∠CBE + x = 180°

⇒ ∠CBE = 180° - x …….(5)

Since, AF is a straight line.

⇒ ∠CDF + ∠ADC = 180°

⇒ ∠CDF + x = 180°

⇒ ∠CDF = 180° - x ……..(6)

From eq.(5) and (6), we get :

⇒ ∠CBE = ∠CDF

In △BEC and △DCF,

⇒ BE = CD (Proved above)

⇒ ∠CBE = ∠CDF (Proved above)

⇒ BC = DF (Proved above)

∴ △BEC ≅ △DCF (By S.A.S axiom)

Hence, proved that △BEC ≅ △DCF.