Question : In $\triangle$ABC, D and E are points on the sides BC and AB, respectively, such that $\angle$ACB = $\angle$ DEB. If AB = 12 cm, BE = 5 cm and BD : CD = 1 : 2, then BC is equal to:
Option 1: $8 \sqrt{3}$ cm
Option 2: $5 \sqrt{5}$ cm
Option 3: $6 \sqrt{5}$ cm
Option 4: $6 \sqrt{3}$ cm
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Correct Answer: $6 \sqrt{5}$ cm
Solution : Given that $\angle ACB = \angle DEB$, we can say that $\triangle ABC$ and $\triangle DBE$ are similar by the Angle-Angle (AA) criterion of similarity. The ratio of corresponding sides in similar triangles is equal. $⇒\frac{AB}{DB} = \frac{BC}{BE}$ Given that AB = 12 cm, BE = 5 cm, and BD : CD = 1 : 2 Since BD : CD = 1 : 2, let $BD = x$ cm and $CD = 2x$ cm $BC = BD + CD = 3x$ cm Substituting these values into the equation, $⇒\frac{12}{x} = \frac{3x}{5}$ $⇒3x^2=60$ $⇒x^2=20$ $⇒x=2\sqrt5\ \text{cm}$ Therefore, $BC = 3x = 6\sqrt5\ \text{cm}$ Hence, the correct answer is $6\sqrt5\ \text{cm}$.
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