In ABC, the midpoint of AB and AC are D and E respectively. CF is parallel to AB and meets DE produced at F. Prove that
(a) DB = FC
(b) DE = EF.
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Answered by
68
Given CF ||AB thus , angle BAC=angleACF...... alternate angles
angleAED=FEC ......by v.o.a
AE=EC.....given E is midpoint of AC
therefore by ASA congruence criteria ,
tri. AED = tri. FEC
by c.s.c.t :
DE=DF.... that's one thing we had to prove...
now , again by c.s.c.t :
AD=FC.....1
but given AD=DB.....d is midpoint..
from 1:
FC=DB
sauveerdixit:
I'll try to figure this out by myself
Anyway, thanks for helping :)
yes, who knows youll find a new way
Yeah
refer triangles lesson in cbse , youll make to the basics
Sure
Finally, I was able solve this one and a couple of questions as well of the same kind!
All credit goes to you ;-)
thanks!
Welcome ^_^
Answered by
24
Answer:
Given CF ||AB thus , angle BAC=angleACF...... alternate angles
angleAED=FEC ......by v.o.a
AE=EC.....given E is midpoint of AC
therefore by ASA congruence criteria ,
tri. AED = tri. FEC
by c.s.c.t :
DE=DF.... that's one thing we had to prove...
now , again by c.s.c.t :
AD=FC.....1
but given AD=DB.....d is midpoint..
from 1:
FC=DB
plz mark as brainliest
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