Question

the medians BD and CE of a triangle ABC meet at G prove that triangle EGD is similar to triangle CGB

Answer 1

Hey there,

draw a line DY || BX.
in triangle BEC,we have
BD/DC=XY/YC     (BY B.P.T THEOREM)
BUT BD=DC
XY/YC=1/1
XY=YC----------------1
SIMILARLY IN TRIANGLE ADC, WE HAVE
AX=XY----------------2

From 1 and 2
AX=XY=YC-------------5

triangle DYC is similar to triangle BXC by AA rule
DY/BX=DC/BC
DY/BE+XE = DC/2DC
DY/BE+XE = 1/2
2DY = BE+XE---------------3

triangle AEX is similar to triangle ADY by AA rule
XE/DY=AX/AY
XE/DY=AX/2AX             (from 5)
XE/DY=1/2
2XE=DY-------------------4

Putting 4 in 3 we get
2(2XE) = BE+XE
4XE=BX+XE
BE=3XE
THEREFORE,
BE/EX = 3XE/XE = 3 : 1

Hope this helps!