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in a trapezoid ABCD, segment AD parallel to segment BC ,E is the midpoint of segment AB , F is the midpoint of segment ,connect point E and point F

Proof : segment EF is parallel to segment , EF=1/2(AD+BC)

Guest Jan 29, 2015

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 #3
avatar+85614 
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From A and C, drop a perpendicular to BD cutting EF at G and H and BD at I and J, respectively.

Then, by similar triangles  AEG = ABI and CFH = CDJ

But, since E anf F are midpoints of AB and CD, and all parts of similar triangles are in the same ratios,  EG = 1/2 of BI and FH = 1/2 of JD. Thus (EG + FH) = 1/2 of (BI + JD).

And AC = GH = IJ    ...thus GH = 1/2(AC + IJ)

Therefore EF = (EG + FH) + GH  =  1/2(BI + JD) + 1/2(AC + IJ) = 1/2(BI + IJ + JD + AC)

But (BI + IJ + JD)  = (BD)

So we have

EF = 1/2(BD + AC) = 1/2(AC + BD)

 

CPhill  Jan 29, 2015
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7+0 Answers

 #1
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Correct me if I'm wrong but I think the segment AD is not parallel to the segment BC... they intersect

Also, I do not mean to be rude or anything like that but what is the question you are asking?

saseflower  Jan 29, 2015
 #2
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I am so sorry. i draw my  picture wrong.Thank you for correct me .

Guest Jan 29, 2015
 #3
avatar+85614 
+5
Best Answer

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

Then, by similar triangles  AEG = ABI and CFH = CDJ

But, since E anf F are midpoints of AB and CD, and all parts of similar triangles are in the same ratios,  EG = 1/2 of BI and FH = 1/2 of JD. Thus (EG + FH) = 1/2 of (BI + JD).

And AC = GH = IJ    ...thus GH = 1/2(AC + IJ)

Therefore EF = (EG + FH) + GH  =  1/2(BI + JD) + 1/2(AC + IJ) = 1/2(BI + IJ + JD + AC)

But (BI + IJ + JD)  = (BD)

So we have

EF = 1/2(BD + AC) = 1/2(AC + BD)

 

CPhill  Jan 29, 2015
 #4
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Okay,guys i am so sorry for typed my oroginal question wrong.

Now i am going to change it to

 

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)

 

i know somebody ignored my new question,so i post it in here.

If you guys mad about i type my original question wrong and  not trying to proof this question, then i will proof thi by myself.

Guest Jan 29, 2015
 #5
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You did good ,Cphill.But i think u misunderstood my given.My given is segment "AC parallel to segment BD"

but that doesnt include "segment EF parallel to segment BD",so angle AEG might not equal to angle ABI.

acctually that are equal ,but why?it neceassary to explain why "EF is parallel to segment BD",then you can proof angle  AEG is eqal to angle ABi.Or u can proof angle  AEG is eqal to angle ABi first , then proof EF is parallel to segment BD.

Guest Jan 29, 2015
 #6
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notcie midpoint

Guest Jan 29, 2015
 #7
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0

so you cant proof triangle AEG is similar to triangle ABI in that way.

try to list the given, remenber proof triangles are similar can either by AA(angle-angle) SAS(side angle side)

SSS(side- side-side).......

Guest Jan 29, 2015

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