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.

Given: A triangle ABC and a line DE parallel to BC.

To prove: A line parallel to one side of a triangle divides the other two sides proportionally.

Proof: Consider ΔABC and DE be the line parallel to Bc, then from ΔABC and ΔADE, we have

∠A=∠A (Common)

∠ADE=∠ABC (Corresponding angles)

Thus, by AA similarity, ΔABC is similar to ΔADE, therefore

AB/AD= AC/AE

⇒AD+DB/AD = AE+EC/AE

⇒1+DB/AD = 1+ EC/AE

⇒DB/AD = EC/AE

Therefore, a line parallel to one side of a triangle divides the other two sides proportionally.

⇒Therefore Proved

Hope this helps!!!

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)

50x5=$250

step-by-step explanation: