CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In given figure through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL=2BL.
1008971_d5584c2f721e4015a4b1b4dab4129206.png

Open in App
Solution

In BMC and EMD, we have

MC=MD [ M is the mid-point of CD]

CMB=EMD [Vertically opposite angles]

and, MBC=MED [Alternate angles]

So, by AAS-criterion of congruence, we have

BMCEMD

BC=DE......(i)

Also, AD=BC [ABCD is a parallelogram].....(ii)

AD+DE=BC+BC

AE=2BC ........(iii)

Now, in AEL and CBL, we have

ALE=CLB [Vertically opposite angles]

EAL=BCL [Alternate angles]

So, by AA-criterion of similarity of triangles, we have

AELCBL

ELBL=AECB

ELBL=2BCBC [Using equation (iii)]

ELBL=2

EL=2BL [Hence proved]

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Angle and Its Measurement
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon