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) ΔDCE ≅ ΔLBE 
(ii) AB = BL.
(iii) AL = 2DC

Answer 1

Given: ABCD is a parallelogram in which is the mid-point of BC.
We need to prove that
(i) ΔDCE ≅ ΔLBE 
(ii) AB = BL.
(iii) AL = 2DC

(i) In  ΔDCE and ΔLBE
∠DCE = ∠EBL   ...[DC || AB, alternate angels]
CE = EB             ...[ E is the midpoint of BC]
∠DEC= ∠LEB    ...[ vertically opposite angels]
∴ By Angel-SIde-Angel Criterion of congruence, we have,
 ΔDCE ≅ ΔLBE
The corresponding parts of the congruent triangles are congruent.
∴ DC= LB        ...[ c. p. c .t]       ....(1)

(ii) DC= AB  ...[ opposite sides of a parallelogram]...(2)
From ( 1 ) and ( 2 ), Ab = BL                      ...(3)

(iii) Al =  AB+ BL                                      ... (4)    
From (3) and (4), Al = AB + AB
⇒AL = 2AB
⇒AL = 2DC                    ...[ From (2) ]