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AGの延長とBCの交点をDとおく
∠GKD= ∠AHD=90°
∠DGK= ∠DAH ( ∵ GK//AH )
∴△DGK ∽ △DAH

AG:GD=2:1
∴DG:DA=1:3
∴GK:AH=1:3

なるほど!ありがとうございます!

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gk:ahは△gbc:△abcの比に等しいのでこれを使います。
重心の定義にそって考えていくと△abc:△gbc=3:1とわかるので、1:3です

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