Question

From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid point of BC.

Prove that :

(i) $\Delta$ DCE $\cong \Delta$ LBE

(ii) AB = BL.

(iii) AL = 2DC

Answer 1

Given: ABCD is a parallelogram. E is the mid point of BC. AB is produced to L.
To prove: (i) ∆DEC ≅ ∆LBE
(ii) AB = BL
(iii) AL = 2DC
Proof: In ∆DCE and ∆LBE
∠DEC = ∠LEB (Vertically opposite angles)
CE = BE (E is the mid point of BC)
∠DCE = ∠EBL (Pair of alternate angles since AB||CD)
∴ ∆DEC ≅ ∆LBE (ASA congruence rule)
CD = BL ...(1) (c.p.c.t)
⇒ AB = CD ...(2) (Opposite sides of parallelogram are equal)
From (1) and (2), we get
AB = BL
AL = AB + BL
⇒ AL = CD + CD (Using (1) and (2))
⇒ AL = 2CD