Question

ABCDE IS A REGULAR PENTAGON.
PROVE THAT :
i)AB║CE
ii)ABEF IS A PARLLELOGRAM

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Answer 1

  Triangle     BDC and CDE are congruent by SAS

∵ CD is common

BC=CD

∠BCD=∠CDE

∴∠CDB=∠DCE

∠DCE=∠DEC (∵CD=DE)

∠A=∠B=∠C=∠D=∠E=108

∠DCE=∠DEC=180-108/2=72/2=36

Similarly, ∠DBC=∠CDB=36

∠BFE=∠CFD=180-∠ECD-∠CDB=180-36-36=108

Thus ∠A=∠F=108......(i)

∠DBA=∠CBA-∠CBD=108-36=72

∠CEA=∠DEA-∠DEC=108-36=72

Thus ∠B=∠E.......(ii)

Hence from (i) and (ii) we find that pairs of opposite angles of quadrilateral ABEF are equal hence it is a parallelogram (actually Rhombus) . This implies further that AB║FEor AB║CE