ABCDE IS A REGULAR PENTAGON.
PROVE THAT :
i)AB║CE
ii)ABEF IS A PARLLELOGRAM
Share
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
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