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Question
In ΔABC, seg DE || side BC. If 2A(ΔADE) = A(⬜ DBCE), find AB : AD and show that BC = `sqrt(3)` DE.
Solution
∴ Given: In `triangleABC`, DE || side BC.
`2A(triangleADE) = A(triangleDBCE)`
To find: `(AB)/(AD)`
To prove: BC = `sqrt3DE`
Proof: In `triangle ABC`
DE | | BC
∴ ∠ADE = ∠ ABC ...Corresponding angle ... (i)
∴ In `triangleADE` and `triangleABC` ∠BAC = ∠DAE ...Common angle
∠ADE = ∠ABC ...By (i)
∴ By A - A test
`triangleADE ≅ triangleABC`
`(AB)/(AD) = (BC)/(DE) = (AC)/(AE)`
Also, `2A(triangleADE) = A(DBCE)`
as `A(triangleABC) = A(triangleADE) +2(DBCE)`
`A(triangleABC) = A(triangleADE) + 2A(triangleADE)`
`A(triangleABC) = 3A(triangleADE)`
`(A(triangleABC))/(A(triangleADE) )= 3/1` ...(2)
`(A(triangleABC))/(A(triangleADE) )= (BC^2)/(DE^2)`
`3/1 = (BC^2)/(DE^2)` ...(Theorem of Area of similar triangle)
`3DE^2 = BC^2`
`sqrt3DE = BC` ...(By taking square root on both sides)
`BC = sqrt3 DE` Hence proved
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