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Question
In figure ABC and DBC are two triangles on the same base BC. Prove that
`"Area (ΔABC)"/"Area (ΔDBC)" = "AO"/"DO"`.
Solution
In ΔAOL and ΔDOM
∠ALO = ∠DMO, ...(90° each)
∠AOL = ∠DOM,
(Vertically opposite 2 sides)
(Vert. opp-angles)
∴ ΔAOL ∼ ΔDOM
∴ `"AL"/"DM" = "AO"/"DO"` ...(1)
If two Δ's are similar the ratio between their corresponding sides is the same.
Now, `"area (ΔABC)"/"area (ΔDBC)" = ((1)/(2) xx "BC" xx "AL")/((1)/(2) xx "BC" xx "Dm") = "AL"/"DM"`
From (1), we get
`"area (ΔABC)"/"area (ΔDBC)" = "AO"/"DO"`.
Hence proved.
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