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

State True or False-
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. ABE is similar to CFB:

A
Always true
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
May be true
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Information insufficient
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Always false
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A Always true

Let EBA=x and EAB=y.
Then, EAB=FCB (Opposite angles of parallelogram)

Also, EBA=CFB (CF||AB )

AEB=CBF

Therefore by AAA Theorem of similarity,

ΔBCF and ΔBAE are similar.


84877_93293_ans.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Criteria for Similarity of Triangles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon