Question

in a triangle ABC the median AD and BE intersect at G.a line DF is drawn parallel to BE such that F is on AC.if AC =9cm,then what is CF equal to?

Answer 1

Given- AC & BE are medians
DF is parallel to BE , AC=9cm
Then,
AE=CE (BE is median on AC therefore E is midpt)

CE = 1\2 of 9cm
CE=4.5 cm
Now
BEC is a triangle
AD is median
therefore D is midpt of BC
and given DF ll BE
therefore by converse of midpt theorem
F is midpt of CE
CF = 1\2 of CE
CF = 2.25cm

Answer 2

Answer:

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