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**tony123****Member**- Registered: 2007-08-03
- Posts: 189

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

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**cberry****Member**- Registered: 2010-02-19
- Posts: 6

Can you post a diagram please.

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**pi_is_exactly3****Member**- Registered: 2010-04-16
- Posts: 3

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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