1. The side AB of a parallelogram ABCD is
produced to X and the bisector of ZCBX
meets DA produced and DC produced at E
and F respectively. Prove that DE = DF =
BA + BC.
Answers
Answer:
ANSWER
Let ∠CBX=x
Given, BF is the bisector of ∠CBX
∴∠CBF=∠XBF=
2
x
∠ABE=∠XBF=
2
x
----Vertically opposite angles
∠EAB=180
∘
−∠DAB ----Sum of angles on a straight line is 180
∘
=180
∘
−x
∠AEB=180
∘
−[∠EAB+∠ABE]=
2
x
In △AEB, ∠ABE=∠AEB=
2
x
----corresponding angles
∴AE=AB
Now we can prove that BC=CF
DE=AD+AE
=AD+AB --------(i)
Similarly,
DF=DC+CF
=DC+BC --------(ii)
AD=BC and AB=DC ---Opposite sides of a parallelogram
So, DE=DF
Hence, DE=DF=AB+BC ---- from (i) and (ii)
solution
Answered By
toppr
660 Views
How satisfied are you with the answer?
This will help us to improve better
answr
Get Instant Solutions, 24x7
No Signup required
girl
More Questions by difficulty
EASY
MEDIUM
HARD
Related Questions to study
The graph of x
2
−4y
2
=0:
MEDIUM
Study later
View Answer
In the adjoining figure, name:
Four pairs of intersecting lines
1777909
MEDIUM
Study later
View Answer
VIEW MORE
Create custom Assignments
Customize assignments and download PDF’s
girl
BROWSE BY
ClassesBoardsExams
MODULES
Online ClassesMock TestsAdaptive PracticeLive Doubts
Stay upto date with our Newsletter!
Stay informed, stay ahead. Be a topper
Your email address
About Us
Brand Resources
Press
Customer Stories
Jobs
Educators
Fellowship
Learning Planet
Guides
Ask
Blog
Bytes
News
Terms of Service
Privacy Policy
Contact Us
FAQs
Google Play
App Store
IN
India