Question

In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD.
Prove that :
(i) ΔABD and ΔECD are congruent. 
(ii) AB = CE.
(iii) AB is parallel to EC

Answer 1

Given: A ΔABC in which D is the mid-point of BC
AD is produced to E so that DE=AD
We need to prove that :

(i) ΔABD and ΔECD are congruent. 

(ii) AB = CE.

(iii) AB is parallel to EC

(i) In ΔABD and ΔECD 

BD = DC       ...[ D is the midpoint of BC ]

∠ADB =∠CDE   ...[ vertically opposite angles ]

AD = DE       ...[ Given ]

∴ By Side-Angle-Side criterion of congruence, we have,

ΔABD ≅ ΔECD

(ii) The corresponding parts of the congruent triangles are congruent.

∴ AB = EC        ...[ c.p.c.t .c] 

(iii) Also, ∠BAD = ∠DEC    ....[ c.p.c t.c ]

∠ABD = ∠DCE            .....[ c.p.c t.c ]

AB || EC    .....[ DAB and DEC are alternate angles ]