Amidsegment triangle abc is parallel to a side of triangle abc what is the missing statement in this proof

Missing part is

Slope of DF = Slope of BC

Reason:Transition property of equality

Step-by-step explanation:

Given:

Proof for a midsegment of triangle ABC is parallel to a side of triangle ABC

In the step by step explanation,

A,B, C are asumed to have vertices (x1,y1)(x2,y2) and (x3,y3)

D and F are defined as mid points.

Coordinates of D and F are found out using mid point formula.

Then slope of BC and slope of DF are found out using two point formula for straight lines.

All these were done correctly.

The missing part is to state that

Slope of DF =Slope of BC (Reason:Transition property of equality)

Next step is done rightly as DF parallel to BC

i need a picture or some numbers

step-by-step explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°