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. This figure does not need to be submitted.

Answer 1

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

Answer 2

The proof that a line that divides two sides of a triangle proportionally is parallel to the third side is E and F coincide on A C. that is, D F coincides with D E. Since D F is parallel to B C, D E is also parallel to B C

Calculations and Parameters

To prove this,

In $ABC$

given, \frac{A D}{D B} = \frac{A F}{E C}........ (1)

   

Let us assume that in $ABC$, the point F is an intersect on the side $AC$.

\frac{A D}{D B} = \frac{A F}{F  C}----- (2)

   

Simplify, in (1) and (2)

\frac{A E}{E C} = \frac{A F}{F  C}

 

Add 1 on both sides,

\frac{A E}{E C} + 1 = \frac{A F}{F  C} +1

\frac{A E + E C}{E C} = \frac{A F + F  C}{F  C}

\frac{A C}{E C}  = \frac{AC}{F  C}

If we apply the Thales Theorem,

E C = F  C

 

From the above, we can say that the points E and F coincide on A C. i.e., D F coincides with D E.

Also, D F is parallel to B C, D E is also parallel to B C

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