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

@maths911 @jinxhead20

Answer 2

@tyree20 @kcooper6 @heybhai @nincompoop

Answer 3

@aleisha96 Do i need to draw figure?

Answer 4

no @heybhai

Answer 5

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.
Refer to the figure below

Answer 6

@heybhai can you help me with one more question please?

Answer 7

Sure @aleisha96

Answer 8

@heybhai Matt is constructing two similar triangles for an art project. The drawing below shows Matt's plans, but there is an error in his drawing. What changes would he make to the dimensions to change the error? Explain your reasoning using complete sentences.

Answer 9

@aleisha96 Matt should just reconstruct the side DF of triangle DEF as 10 units instead of 10.5 units for similarity as the ratio of side must remain equal to 0.8