Question

Medal:
Prove that a line that divides two sides of a triangle proportionally is parallel to the third side. Be sure to create and name the appropriate geometric figures. This figure does not need to be submitted.

Answer 1

@robtobey please help

Answer 2

@wolf1728 urgent help please!!!

Answer 3

Given
In ΔABC, D and E are the two points of AB and AC respectively,
such that, AD/DB = AE/EC.

To Prove
DE || BC

Proof
In ΔABC,
given, AD/DB = AE/EC ----- (1)

Let us assume that in ΔABC, the point F is an intersect on the side AC.
So we can apply the Thales Theorem, AD/DB = AF/FC ----- (2)

Simplify, in (1) and (2) ==> AE/EC = AF/FC
Add 1 on both sides, ==> (AE/EC) + 1 = (AF/FC) + 1
==> (AE+EC)/EC = (AF+FC)/FC
==> AC/EC = AC/FC
==> EC = FC
From the above, we can say that the points E and F coincide on AC.
i.e., DF coincides with DE.
Since DF is parallel to BC, DE is also parallel BC
This is also know as Converse of Basic Proportionality theorem is proved.
Refer to the figure below

Answer 4

yeah i dont understand yet

Answer 5

such slow load!

Answer 6

That is the best that I can explain...Sorry...

Answer 7

k thanks

Answer 8

@ranga

Answer 9

were did @YanaSidlinskiy get 'F' from? There isn't even an F on the triangle