In a triangle ABC, D is mid-point of BC; AD is produced upto E, so that DE = AD. Prove that :

(i) △ ABD and △ ECD are congruent.

(ii) AB = EC

(iii) AB is parallel to EC.

△ ABC with AD produced upto E is shown in the figure below:

(i) In △ ABD and △ ECD,

⇒ AD = DE (Given)

⇒ BD = DC (As D is the mid-point of BC)

⇒ ∠ADB = ∠CDE (Vertically opposite angles are equal)

∴ △ ABD ≅ △ ECD (By S.A.S. axiom)

Hence, proved that △ ABD ≅ △ ECD.

(ii) Since,

△ ABD ≅ △ ECD (Proved above)

We know that,

Corresponding parts of congruent triangles are equal.

∴ AB = EC.

Hence, proved that AB = EC.

(iii) From figure,

∠ABD = ∠DCE (By C.P.C.T.C.)

Since, these are alternate angles and are also equal,

thus, it can be said, AB and EC are parallel with AE as transversal.

Hence, proved that AB is parallel to EC.