# is there converse for side splitter theorem?

• Sep 18th 2012, 04:56 AM
mido22
is there converse for side splitter theorem?[SOLVED]
I know that :-

if a line is parallel to a side of a triangle and intersect the other two sides, then this line divides those two sides proportionally.

but
if i have that the line divides the two sides proportionally can i get that this line is parallel to the 3rd side
http://www.mathwarehouse.com/geometr...er-theorem.gif
if i have AB:BD=AC:CE=1:4
is it true to say that BC // DE
• Sep 18th 2012, 04:59 AM
HallsofIvy
Re: is there converse for side splitter theorem?
Yes, and it is easy to prove. Because of the proportions, the triangles ABC and ADE are "similar" so have congruent angles. The by "corresponding angles" the lines are parallel.
• Sep 18th 2012, 05:33 AM
mido22
Re: is there converse for side splitter theorem?
but i have only 2 proportional sides AB:BD=AC:CE=1:4 only
• Sep 18th 2012, 06:26 AM
emakarov
Re: is there converse for side splitter theorem?
Quote:

Originally Posted by mido22
but i have only 2 proportional sides AB:BD=AC:CE=1:4 only

Use condition #3 (two sides in the same ratio with the same angle between them) here.
• Sep 18th 2012, 03:00 PM
mido22
Re: is there converse for side splitter theorem?
thNX very much