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Question
Prove that the area of the triangle BCE described on one side BC of a square ABCD as base is one half of the area of similar triangle ACF described on the diagonal AC as base.
Solution
ABCD is a square. ∆BCE is described on side BC is similar to ∆ACF described on diagonal AC.
Since ABCD is a square. Therefore,
AB = BC = CD = DA
and AC = `sqrt(2)"BC"` ...[∵ Diagonal = `sqrt(2)` (side)]
Now, ∆BCE ~ ∆ACF
⇒ `"Area (∆BCE)"/"Area (∆ACF)" = "BC"^2/"AC"^2`
⇒ `"Area (∆BCE)"/"Area (∆ACF)" = "BF"^2/(sqrt(2)"BC")^2 = (1)/(2)`
⇒ Area (∆BCE) = `(1)/(2) "area (∆ACF)"`.
Hence proved.
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