Question

9. 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) triangle DCE is congruent to triangle LBE
(ii) AB = BL.
(iii) AL = 2DC​

Answer 1

hello!!

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