In a triangle d and e are two points on sides AB and AC and de parallel to BC if bd =x-3,AB =2x ,ce =x-2 and AC =2x+3.Find x
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DE ll BC
angle D = angle B
angle A = angle A
= ∆ABC ~ ∆ADE (AA similarity)
== AD/AB = AE/ AC
= AB-BD / AB = AC-EC/ AC
= 2x-(x-3) / 2x = 2x+3-(x-2)
= 2x-x+3/2x = 2x+3-x+2/2x+3
= x+3/ 2x = x+5/ 2x+3
= 2x2 + 6x +3x +9 = 2x2 +10x
= 9x +9 = 10x
= x =9
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