# Prove that G divides AC into two equal segments, if CA is paralel to ED?

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- PopeLv 72 years agoFavorite Answer
This is going to be somewhat abbreviated. First prove these similarity statements:

∆EHF ~ ∆AGF

∆DHF ~ ∆CGF

Therefore, CG/DH = AG/EH = GF/HF

∆BHE ~ ∆BGC

∆BHD ~ ∆BGA

Therefore, CG/EH = AG/DH = BG/BH

CG/DH = AG/EH and CG/EH = AG/DH

(CG/DH)(CG/EH) = (AG/EH)(AG/DH)

CG²/[(DH)(EH)] = AG²/[(DH)(EH)]

CG² = AG²

CG = AG

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