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Question
In the given figure, DE║BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.
Solution
In ΔADE and ΔABC
∠𝐴𝐷𝐸 = ∠𝐴𝐵𝐶 (𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑛 𝐷𝐸 ∥ 𝐵𝐶)
∠𝐴𝐸𝐷 = ∠𝐴𝐶𝐵 (𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑛 𝐷𝐸 ∥ 𝐵𝐶
By AA similarity criterion, ΔADE ~ ΔABC
If two triangles are similar, then the ratio of their corresponding sides are proportional
∴ `(AD)/(AB)=(AE)/(AC)`
⇒`( AD)/(AD+DB)=(AE)/(AE+EC)`
⇒` x/(x+3x+4)=(x+3)/(x+3+3x+19)`
⇒ `x/(4x+4)=(x+3)/(x+3+3x+19)`
⇒ `x/(2x+2)=(x+3)/(2x+11)`
⇒ `2x^2+11x=2x^2+2x+6x+6`
⇒ 3x=6
⇒ x=2
Hence, the value of x is 2.
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