Question

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.

Answer 1

i need help :(

Answer 2

@bibby

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.

Answer 4

omg finally thank you!!

Answer 5

Haha Your Welcome

Answer 6

i have one last question on my quiz can u also help with that too?

Answer 7

yeah I'll try