In the given figure, ABCD is a square and EF is parallel to diagonal BD. if EM=FM, then prove that:a)DF=BE b)AM bisects angleBAD
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See diagram.
AB = BC = CD = DA.
ΔCFE and ΔCDB are similar as their corresponding sides CF || CD, CE || CB and EF || BD (given).
CE / CB = CF / CD = FM / EM
=> CE = CF ---(1)
BE = BC - CE
DF = DC - CF
= BC - CE by (1)
=> DF = BE
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(ii)
Draw perpendiculars from M onto AB, BC, CE and DA.
ΔCMF and ΔCME are similar and congruent, as : CE = CF, CM is common, and FM = EM. Then the altitudes from M, MI = MH.
Hence, MG = GI - MI = BC - MI
MJ = HJ - MH = CD - MH
=> MG = MJ
As G and J are on the sides of ∠BAD, and MG = MJ, M lies on the angular bisector of ∠BAD.
AB = BC = CD = DA.
ΔCFE and ΔCDB are similar as their corresponding sides CF || CD, CE || CB and EF || BD (given).
CE / CB = CF / CD = FM / EM
=> CE = CF ---(1)
BE = BC - CE
DF = DC - CF
= BC - CE by (1)
=> DF = BE
==============
(ii)
Draw perpendiculars from M onto AB, BC, CE and DA.
ΔCMF and ΔCME are similar and congruent, as : CE = CF, CM is common, and FM = EM. Then the altitudes from M, MI = MH.
Hence, MG = GI - MI = BC - MI
MJ = HJ - MH = CD - MH
=> MG = MJ
As G and J are on the sides of ∠BAD, and MG = MJ, M lies on the angular bisector of ∠BAD.
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