If abc is an equilateral triangle with ad perpendicular bc prove ad²=3dc²
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given:AB=BC=CD
AD⊥BC
to prove:AD²=3DC²
proof:in ΔABD
AB²=AD²+BD²
AD²=AB²-BD²
AD²=BC²-BD²
AD²=(2BD)²-BD²
AD²=4BD²-BD²
AD²=3BD²
AD²=3DC²
AD⊥BC
to prove:AD²=3DC²
proof:in ΔABD
AB²=AD²+BD²
AD²=AB²-BD²
AD²=BC²-BD²
AD²=(2BD)²-BD²
AD²=4BD²-BD²
AD²=3BD²
AD²=3DC²
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