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#1 2008-03-20 06:07:05

tony123
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convex pentagon

Let ABCDE be a convex pentagon such that BC = CD = DE
and each diagonal of the pentagon is parallel to one of its sides. Prove that
all the angles in the pentagon are equal, and that all sides are equal

 

#2 2010-02-22 21:26:30

cberry
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Re: convex pentagon

Can you post a diagram please.

 

#3 2010-04-20 04:20:43

pi_is_exactly3
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Re: convex pentagon

ok so this is just tons and tons of simple steps but here it goes

first by drawing the pentagon and since parrallel lines connecting the same points are the same line we see that the sets of parrallel lines are

BC DA     DC EB        ED AC           AE BD           AB EC

then by using both iscosoles triangle and parrallel line angle theorems (Z patterns) be see

1)DEC=DCE=BCA=CAD=ADE=ECA=CAB let equal x
2)CBD=CDB=BDA=DAE=DBE=BEA let equal y
3) and finally ABE=BEC  let equal z

so recap we now have angles (pentagon) which equal

ABC= z+2y   BCD=3x  CDE=x+2y DEA=x+y+z  EAB=2x+y

By looking further at inner triangles
for ADE we see 2y+2x+z=180
and for CDE we see 3x+2y=180  this implies that z=x

Triangle ABC has contains 2 equal angles BAC and BCA which implies iscoscoles and therefore sides
AB=BC  which from above also equal DC=DE
this is sufficient enough to say the pentagon is regular

 

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