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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
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Solution
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) ]
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