Question

In the given figure, ABCD is a parallelogram, E is the mid-point
of BC. DE produced meets AB produced at L.
Prove that:
(i) AB - BL
(ii) AL = 2 DC.
​

Answer 1

Answer:

ABCD is a parallelogram

so

AB=DC, AD=BC

if one of the parallel lines is extended, the two lines will still be parallel

therefore, DC parallel to BL

in triangle BEL and DEC

BE=EC(E midpoint)

angle ELB = angle EDC  ( alternate interior angles)

angle EBL = angle ECB  ( alternate interior angles)

therefore triangle BEL and DEC are congruent by AAS congruency rule

BL=DC (by CPCT)-------(1)

AB=DC

AB=BL(using (1) )

hence proved ( i )

AL=AB+BL

AL=AB+AB (using ( 1 )

AL=DC+DC (given in question)

therefore,

AL=2 DC

HENCE PROVED

hope it helps : )

Answer 2

Step-by-step explanation:

Given,

ABCD is a Parallelogram

so,

AB = BL

AD = BC

DC || BL

To Prove,

(i) AB = BL

(ii) AL = 2 DC.

Proof,

(i) AB = BL

∆ BEL = ∆ DEC

BE = EC [ given]

angle BLE = angle EDC [ alternate angles]

angle DCE = angle LBE [ alternate angles]

∆ BEL = ∆ DEC by AAS.

AB = BL [ c. p. c. t] —(1)

(ii) AL = 2 DC

AB = BL [ using (i)

so,

AL = AB + BL

AL = AB + AB [ using (i)

AL = AD + AD [ given in question]

AL = 2 DC

therefore, AL = 2DC