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One can show that BD⊥EF. Then S(BEDF)= 1/2×BD×EF= BO×EF= S(ABC) (by Problem 1055)Thus sobtracting S(BEGMF) both side, S₁+S₂=S₃
Join DA, DC,DH. Let (XYZ) stand for area of ΔXYZ.AD // EH implies (AED) = (AHD). Follows S1 = (GHD).DC // HF implies (DFC) = (DHC). Follows S2 = (MHD).Hence S1 + S2 = S3.
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One can show that BD⊥EF.
ReplyDeleteThen
S(BEDF)
= 1/2×BD×EF
= BO×EF
= S(ABC) (by Problem 1055)
Thus sobtracting S(BEGMF) both side,
S₁+S₂=S₃
Join DA, DC,DH. Let (XYZ) stand for area of ΔXYZ.
ReplyDeleteAD // EH implies (AED) = (AHD). Follows S1 = (GHD).
DC // HF implies (DFC) = (DHC). Follows S2 = (MHD).
Hence S1 + S2 = S3.