Can this be used as a proof? (1 Viewer)

[nightweaver066]

If i had this,

Can i prove that those two lines are parallel simply by using the converse of ratio of intercepts on parallel lines? Or do i have to prove similarity and then prove that they are parallel?

Thanks

 

[qawe]

neither of these. you use this: "the line joining the midpoints of two sides of a triangle is parallel to the 3rd side (and half its length)" - its on the syllabus

 

[nightweaver066]

neither of these. you use this: "the line joining the midpoints of two sides of a triangle is parallel to the 3rd side (and half its length)" - its on the syllabus

It's not joining the midpoints but cutting the two equal sides of an isosceles triangle equally..

 

[thoth1]

neither of these. you use this: "the line joining the midpoints of two sides of a triangle is parallel to the 3rd side (and half its length)" - its on the syllabus

i neva new bout this. tnx

 

[Drongoski]

For 2U maths, yr approach not quite adequate. It is better to:

i) show similarity of 2 triangles by virtue of proportionality of pairs of sides & equality of included angles

ii) then use equality of a pair of corresponding angles to conclude 2 relevant lines are parallel

 

[taeyang]

For 2U maths, yr approach not quite adequate. It is better to:

i) show similarity of 2 triangles by virtue of proportionality of pairs of sides & equality of included angles

ii) then use equality of a pair of corresponding angles to conclude 2 relevant lines are parallel

^This

 

[nightweaver066]

For 2U maths, yr approach not quite adequate. It is better to:

i) show similarity of 2 triangles by virtue of proportionality of pairs of sides & equality of included angles

ii) then use equality of a pair of corresponding angles to conclude 2 relevant lines are parallel

What do you mean for 2U maths?

Do you mean this is not a really suitable way to explain it this way in 2U math but possible in 3U and 4U math?

 

[Drongoski]

In this case,(I shouldn't have qualfied it with the 2U), you are almost assuming the very proposition you are trying to prove.

 

[MrBrightside]

If your given the points, you can use gradient formula to find the gradient of both lines then it's just m1 = m2

 

[hscishard]

I've seen an answer using the converse. Forgot which book it was from.
Drongoski's method is fail-safe and I would've done that for sure

 

[Drongoski]

If your given the points, you can use gradient formula to find the gradient of both lines then it's just m1 = m2

In Euclidean geometry we don't use gradients. Even if we can, where are we told or how do we know m1 = m2?