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in a trapezoid ABDC, segment AC parallel to segment BD ,E is the midpoint of segment AC , F is the midpoint of segment CD,connect point E and point F

Proof : segment EF is parallel to segment BD ,      EF=1/2(AC+BD)

Here is my proof :

first connect point F and point A, extend segment AF to point K, extend segment BD and intercept segment AK at point K

Because point F is the midpoint of segment DC

so DF=CF

becuase AC is parallel to BD ,BC extend to point K

so AC is parallel to BK

Therefore,angle FAC= angle FKD, angle FCA =angle FDK (two alternate interior angles are equal in two parallel lines)

so ,triangle ACF is congruent to triangle KDF(AAS)

so AC=KD ,AF=KF (corresponding sides of congruent triangles are equal)

so,AF/AK=AF/(AF+KF)=AF/(AF+AF)=AF/2AF=1/2

because point E is the midpoint of AB

so, AE=BE

same,AE/AB=1/2

so, AE/AB=AF/AK

and because angle EAF = angle BAK

so triangle EAF is similar to triangle BAK

so, EF/BK=AE/AB=AF/AK=1/2

so, EF=1/2 BK= 1/2 (BD+DK)

because DK= AC

EF= 1/2 (BD+AC)

because triangle EAF is similar to triangle BAK

so, angle AEF= angle ABK

so, EF is parallel to BC

 Jan 30, 2015

Best Answer 

 #4
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Acctually,i had learned how to prove this two years ago.But now i forgot some,so i decided to prove this again.

and post this post on the forum.

i  enjoy at prove any math theory or law.so yeah....................

 Jan 30, 2015
 #2
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+5

From A and C, drop a perpendicular to BD cutting EF at G and H and BD at I and J, By CPHill

and i am going draw it out right now 

my given is "in a trapezoid ABDC, segment AC parallel to segment BD ,E is the midpoint of segment AC , F is the midpoint of segment CD,connect point E and point F"

so EF is the median line of trapezoid ABDC , but is EF is parallel to BD?it does, but in this case we have to proof it.

if we cant proof EF is parallel to BD, then we cant get angle AEG= angle ABI or angle AGE = angle AIB.

so we cant proof triangle AEG is similar to triangle ABI is this way.

on the other hand , If proof two triangles are similar ,we need SSS (side-side-side),SAS (Side -side- side)or AA (angle angle).But we only have angle AEG =angle ABI (angle) , and AE=BE (side).so we do not have enough to proof triangle AEG is similar to triangle ABI......

careful with my given guys.

 Jan 30, 2015
 #3
avatar+118608 
+5

I am really sorry that we did not get to your question we juust have too many questions to get through them all but it is really good that you did it yourself.   

Are you happy with you answer?

 Jan 30, 2015
 #4
avatar
+5
Best Answer

Acctually,i had learned how to prove this two years ago.But now i forgot some,so i decided to prove this again.

and post this post on the forum.

i  enjoy at prove any math theory or law.so yeah....................

Guest Jan 30, 2015
 #5
avatar+118608 
0

Okay, that is really good.  I like working through proofs too,  they can be very challenging  

 Jan 30, 2015

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