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Let ABCDE be a convex pentagon such that BC = CD = DEand each diagonal of the pentagon is parallel to one of its sides. Prove thatall the angles in the pentagon are equal, and that all sides are equal
Can you post a diagram please.
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 ECthen by using both iscosoles triangle and parrallel line angle theorems (Z patterns) be see 1)DEC=DCE=BCA=CAD=ADE=ECA=CAB let equal x2)CBD=CDB=BDA=DAE=DBE=BEA let equal y3) and finally ABE=BEC let equal zso recap we now have angles (pentagon) which equalABC= z+2y BCD=3x CDE=x+2y DEA=x+y+z EAB=2x+yBy looking further at inner trianglesfor ADE we see 2y+2x+z=180and for CDE we see 3x+2y=180 this implies that z=xTriangle 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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