# Trapezium Proof

• January 16th 2011, 03:11 PM
jacs
[SOLVED with many thanks] Trapezium Proof
ABCD is a trapezium, with AB||CD. A is joined to C and B is joined to D. Point E is the midpoint of BD and point F is the midpoint of AC. E is joined to F.
Show that the line EF is half the distance of the difference between DC and AB.

Not sure if this is sound but was how I started:
If E and and F are midpoints of BD and AC respectively, and AB||CD then EF must be parallel to AB and CD (family of parallel lines cut off intercepts in equal ratios)

then i thought that I could use similar triangles. Letting diagonals intersect at X and after proving $\Delta$XEF similar $\Delta$XDC and $\Delta$AXB with sides in equal ratios, but can't figure out how to get it structured right or proceed from there.

Been at it a while now and it perhaps has become a issue of can't see the forest for the trees, but am defnintely hittng a brick wall.

any help greatly appreciated
• January 16th 2011, 04:01 PM
Quote:

Originally Posted by jacs
ABCD is a trapezium, with AB||CD. A is joined to C and B is joined to D. Point E is the midpoint of BD and point F is the midpoint of AC. E is joined to F.
Show that the line EF is half the distance of the difference between DC and AB.

Not sure if this is sound but was how I started:
If E and and F are midpoints of BD and AC respectively, and AB||CD then EF must be parallel to AB and CD (family of parallel lines cut off intercepts in equal ratios)

then i thought that I could use similar triangles. Letting diagonals intersect at X and after proving $\Delta$XEF similar $\Delta$XDC and $\Delta$AXB with sides in equal ratios, but can't figure out how to get it structured right or proceed from there.

Been at it a while now and it perhaps has become a issue of can't see the forest for the trees, but am defnintely hittng a brick wall.

any help greatly appreciated

Extend the line [AB] out to the right and left, forming a rectangle.

Then extend [EF] until the line meets [BC] at G.

Then it is easy to see that |EG|=0.5|DC| and |FG|=0.5 |AB|

from which the result follows.

(There's really no need to draw the rectangle, it may help make the picture a little clearer).
• January 16th 2011, 04:15 PM
abhishekkgp
Quote:

Originally Posted by jacs
ABCD is a trapezium, with AB||CD. A is joined to C and B is joined to D. Point E is the midpoint of BD and point F is the midpoint of AC. E is joined to F.
Show that the line EF is half the distance of the difference between DC and AB.

Not sure if this is sound but was how I started:
If E and and F are midpoints of BD and AC respectively, and AB||CD then EF must be parallel to AB and CD (family of parallel lines cut off intercepts in equal ratios)

then i thought that I could use similar triangles. Letting diagonals intersect at X and after proving $\Delta$XEF similar $\Delta$XDC and $\Delta$AXB with sides in equal ratios, but can't figure out how to get it structured right or proceed from there.

Been at it a while now and it perhaps has become a issue of can't see the forest for the trees, but am defnintely hittng a brick wall.

any help greatly appreciated

join AE and let AE meet DC at G.
then triangles AEB and DEF are congruent. Hence AE=EG.
Then in triangle AGC observe that E is mid-point of AG and F is mid-point of AC. now use mid-point theorem(or triangle mid-line theorem) to get EF=(1/2)*(DC-DG)
but DG=AB hence EF=(1/2)(DC-AB).
• January 16th 2011, 04:18 PM
Plato
Attachment 20464Look at the extended diagram .
Note that $2m(\overline{PQ})= m(\overline{AB})+ m(\overline{CD})$.http://www.mathhelpforum.com/math-he...4&d=1295226960

Also note that $2m(\overline{FP})= m(\overline{DC})$.

If we continue this way, then the result will follow.
• January 16th 2011, 05:13 PM
jacs
Thank you all so very much, this has been a massive help and it is awesome to see so many different approaches. This has been incredibly helpful and has given me a few ideas to think 'outside the box' or trapezium, in this case. Really appreciate your help!

Thanks again