Question

Look at the figure shown below.



Which statement must be true to prove that segment DE is parallel to segment CA?

http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/pool_Geom_3641_1008_08/image0044e8ca51a.jpg

1 : DC = 5 : EA

4 : DC = 3 : EA

1 : EA = 5 : DC

4 : EA = 3 : DC

Answer 1

@Callisto

Answer 2

HELP

Answer 3

@rebeccaskell94

Answer 4

@panlac01 lol

Answer 5

if CD and EA are congruent, then DE and CA are parallel

Answer 6

i think its choice two because it will end up being 60 and 60

Answer 7

that or choice D

Answer 8

do you see that BD and BE are not congruent?

Answer 9

yeah

Answer 10

I just saw it... lol

Answer 11

forget what I said earlier

Answer 12

Use intercept theorem...
\[\frac{BD}{DC}=\frac{BE}{EA}\]

Answer 13

so B :)

Answer 14

Put the values into it, and simplify

Answer 15

B B B B

Answer 16

That's... what I think.. only....
Btw, it's not intercept theorem... /_\

Answer 17

it is not

Answer 18

@panlac01 Do you have any idea??

Answer 19

yeah. Triangle mid-segment

Answer 20

nvm

Answer 21

@panlac01 May I know if the ratio I've written is correct??

Answer 22

4:3

Answer 23

triangle inequality theorem

Answer 24

maybe it is about time to take back my medals :-)

Answer 25

either BD and CD have to be congruent or BE and EA

Answer 26

that's following the median of triangle

Answer 27

I was thinking if \[\frac{AX}{XB} = \frac{AY}{YC}\], it's more like XY//BC. Something similar to the mid-point theorem, but it's not mid-point theorem...