I need geometry help

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

Take a triangle ABC with vertex at A. Mark the midpoint of side AB with point D and the midpoint of side BC with point E. Join the midpoints to form side DE. Now you have two triangles ABC and ADE where triangle ADE is included in triangle ABC. Prove that DE is parallel to BC.

Prove that triangles ABC and ADE are similar.

Proof:

1- The angle at vertex A is common for both ABC and ADE. Congruent angles

2- by law of proportionality : AC/AE = AB/AD = 2 , and therefore proportional sides are congruent

such that AB is congruent to AD and AC is congruent to AE

With one angle and two sides congruent, the SSA postulate proves that triangles ABC and ADE are similar. Now similar triangles has the same angles and angles at D and E are equal to angles at B and C respectively. the angles are also corresponding to each other such that angle D is corresponding to angle B and angle E is corresponding to angle C. Therefore according to the law of corresponding angles side DE is parallel to side BC

Answer 2

Triangle ABC: A on top, B on the left, C on the right.

line MN divides the triangle proportionally so that is APPEARS to be parallel to side BC.

(It shall be proven that MN is in fact parallel)

x = AM, y = AN, z = MB, and w = NC.

AM and AB are proportional

AN and AC are proportional

angle MAN and BAC are equal by reflexive.

triangle MAN and triangle BAC are similar by SAS

angle AMN = angle ABC by similarity/corresponding parts

angle ANM = angle ACB by similarity/corresponding parts

since the corresponding angles are congruent, the lines are parallel