Geometry problem midpoint theorem

[mathlearn]

A problem on geometry proof

Hi (Smile),

View attachment 5807

When considering the \(\displaystyle \triangle\) ABM E is the midpoint of AB
& EO //OM (given).I think this is the way to tell AO=OM , Help .Many Thanks (Smile)

 

[mrtwhs]

There are a lot of details to consider but the key fact needed is \(\displaystyle \triangle ODC \cong \triangle MDB\) by SAA.

 

[mathlearn]

Many Thanks (Smile)

But I don't think there's a need to do so :)

I Think It has got something to do with

When considering the \(\displaystyle \triangle ABM\) E is the midpoint of AB
& EO //OM (given)

Can you figure it out?Help.

Many Thanks (Smile)

 

[mrtwhs]

Many Thanks (Smile)

But I don't think there's a need to do so :)

I Think It has got something to do with

When considering the \(\displaystyle \triangle ABM\) E is the midpoint of AB
& EO //OM (given)

Can you figure it out?Help.

Many Thanks (Smile)

I didn't mean to imply that my approach was the only way to do the problem. I just thought it would be straightforward. After you get the triangles I mentioned congruent, it is trivial that BMCO is a parallelogram and the rest of the problem follows.

The theorem that you are thinking of is a standard high school result:

In a plane, if a line goes through the midpoint of one side of a triangle and is parallel to a second side of the triangle, then it goes through the midpoint of the third side of the triangle.

 

[mathlearn]

A nice approach by saying the two triangles \(\displaystyle ODC \cong MBA\) :)

By considering \(\displaystyle \triangle ABM\)

It is given that 'E' is the midpoint of AB. According to mrtwhs

"In a plane, if a line goes through the midpoint of one side of a triangle and is parallel to a second side of the triangle, then it goes through the midpoint of the third side of the triangle"

Line EO meets the side AM of the \(\displaystyle \triangle ABM\)

and it is also given that EC//BM \(\displaystyle \therefore EO//BM\)

And using the converse of the midpoint theorem

The straight line through the midpoint of one side of a triangle and parallel to another side,bisects the third side.

It can be proved that AO=OM;(Smile)

And can you help me tell that MC//BF and 2AD=3AO

Many Thanks :)

 

[Greg]

i) and ii) can be proved using the midpoint theorem. Using the fact that BM || EC and MC || BF, show that triangles BOC, BMC are congruent and triangles CMO and BMO are congruent. From that it follows that BMCO is a parallelogram (each diagonal of a parallelogram divides the parallelogram into two congruent triangles). To prove iv), use the fact that the diagonals of a parallelogram bisect each other.

 

[mathlearn]

Many Thanks :)

When considering the \(\displaystyle \triangle AMC\)

O is the midpoint of AM (AO=OM)

F is the midpoint of AC (AF=FC)

The straight line through the midpoint of one side of a triangle and parallel to another side,bisects the third side.

\(\displaystyle \therefore OF//MC \)

\(\displaystyle \therefore MC//BF\)

And I now see

View attachment 5814

Help me to proceed

Many Thanks :)

 

[Evgeny.Makarov]

Concerning $2AD=3AO$: We have proved that $AO=OM$ and that $BMCO$ is a parallelogram. Diagonals in a parallelogram are divided in half; therefore, $OD=DM$. Thus, $2AD=2(AO+OD)=2(OM+OD)=2(2OD+OD)=6OD$ and $3AO=3OM=3\cdot2OD=6OD$.

 

[mathlearn]

Thank you very much ! (Smile)