Question

17. 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.

How to solve this???

Answer 1

Answer:

Step-by-step explanation:

given: ABCD is a parallelogram. E is the midpoint of AD and DL is parallel to EB. meets AB produced at F.

TPT: B is the midpoint of AF and EB = LF

proof:

in triangle ADF , 

E is the midpoint of AD and EBDF 

therefore by the converse of mid point thm.

 B is the mid point of AF.

in triangle ADF,

E and B are the mid points AD and AF respectively

by the mid point thm EB = 1/2 DF.........(1)

since EB is parallel to DL and BL is parallel to ED.

EBLD is a parallelogram .

EB =DL......(2) [opposite sides of a parallelogram are equal]

DL = 1/2 DF i.e. L is the midpoint of DF.

LF = DL = EB

therefore EB = LF

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