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Question
In ΔABC; BM ⊥ AC and CN ⊥ AB; show that:
`(AB)/(AC) = (BM)/(CN) = (AM)/(AN)`
Solution
In ΔABM and ΔACN,
∠AMB = ∠ANC ...(BM ⊥ AC and CN ⊥ AB)
∠BAM = ∠CAN ...(Common angle)
`=>` ΔABM ∼ ΔACN ...(AA criterion for similarity)
`=> (AB)/(AC) = (BM)/(CN) = (AM)/(AN)`
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