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Question
In the given figure, DE || BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.
- Write all possible pairs of similar triangles.
- Find lengths of ME and DM.
Solution
i. In ΔAME and ΔANC,
∠AME = ∠ANC ...(Since DE || BC that is, ME || NC)
∠MAE = ∠NAC ...(Common angle)
`=>` ΔAME and ΔANC ...(AA criterion for similarity)
In ΔADM and ΔABN,
∠ADM = ∠ABN ...(Since DE || BC that is, ME || BN)
∠DAM = ∠BAN ...(Common angle)
`=>` ΔADM and ΔABN ...(AA criterion for similarity)
In ΔADE and ΔABC,
∠ADE = ∠ABC ...(Since DE || BC that is, ME || NC)
∠AED = ∠ACB ...(Since DE || BC)
`=>` ΔADE and ΔABC ...(AA criterion for similarity)
ii. In ΔAME and ΔANC,
∠AME = ∠ANC ...(Since DE || BC that is, ME || NC)
∠MAE = ∠NAC ...(Common angle)
`=>` ΔAME and ΔANC ...(AA criterion for similarity)
`=> (ME)/(NC) = (AE)/(AC)`
`=> (ME)/6 = 15/24`
`=>` ME = 3.75 cm
In ΔADE and ΔABC,
∠ADE = ∠ABC ...(Since DE || BC that is, ME || NC)
∠AED = ∠ACB ...(Since DE || BC)
`=>` ΔADE and ΔABC ...(AA criterion for similarity)
`=> (AD)/(AB) = (AE)/(AC) = 15/24` ...(i)
In ΔADM and ΔABN,
∠ADM = ∠ABN ...(Since DE || BC that is, ME || NC)
∠DAM = ∠BAN ...(Common angle)
`=>` ΔADM and ΔABN ...(AA criterion for similarity)
`=> (DM)/(BN) = (AD)/(AB) = 15/24` ...(From (i))
`=> (DM)/24 = 15/24`
`=>` DM = 15 cm
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