SSC CHSL 10 Jan 2017 Shift 3 Paper

Given : AB is parallel to CD. AE = 10 cm , BC = 15 cm and ED = 15 cm
To find : EC = ?
Solution : Let EC = x cm
In ∆BAE and ∆CDE
⇒ ∠ABE = ∠DCE (Alternate interior angles)
⇒ ∠BAE = ∠CDE (Alternate interior angles)
⇒ ∠AEB = ∠CED (Vertically opposite angles)
Thus, ∆BAE ∼ ∆CDE (AAA criteria)
∴ $\frac{\mathit{A}\mathit{E}}{\mathit{D}\mathit{E}}=\frac{\mathit{E}\mathit{B}}{\mathit{E}\mathit{C}}$
⇒ $\frac{10}{15}=\frac{\left(15-\mathit{x}\right)}{\mathit{x}}$
⇒ $\frac{\left(15-\mathit{x}\right)}{\mathit{X}}=\frac{2}{3}$
⇒ $45-3\mathit{x}=2\mathit{x}$
⇒ $2\mathit{x}+3\mathit{x}=5\mathit{x}=45$
⇒ $\mathit{x}=\frac{45}{5}=9$ cm