Prove DE//BC, given mBD=mDC, F is on median AD, Lines CD, BE meet AD at F.

[yma16]

Given mBD=mDC, F is a point on the median AD. Lines CD and BE meet AD at F.

 

[Steven G]

View attachment 9610Given mBD=mDC, F is a point on the median AD. Lines CD and BE meet AD at F.

Hi, you have beed around here long enough to know that to get help you need to show us your work. Thanks!

 

[Dr.Peterson]

View attachment 9610Given mBD=mDC, F is a point on the median AD. Lines CD and BE meet AD at F.

If you're still having trouble with this, you might try constructing E' on AC such that DE' || BC, and see if you can show that E=E'. That sort of thinking is needed for some tricky problems like this.

If this doesn't help, tell us the context -- was this a problem in a chapter talking about some particular theorems that might be of use, for example?

 

[yma16]

I thought about this and did not find it was helpful.

If you're still having trouble with this, you might try constructing E' on AC such that DE' || BC, and see if you can show that E=E'. That sort of thinking is needed for some tricky problems like this.

If this doesn't help, tell us the context -- was this a problem in a chapter talking about some particular theorems that might be of use, for example?

I proved it by adding a line //AB and pass C. I have not got time to write it neatly and post it hear.

 

[Dr.Peterson]

I proved it by adding a line //AB and pass C. I have not got time to write it neatly and post it hear.

Sounds good.

 

[yma16]

proof

Sounds good.

This is the proof.

 

[Dr.Peterson]

This is the proof.

Looks good, too.

I probably would have done something very different, having a different background, but this definitely works.