Problem of the Week

Problem 14 Solution

$\textbf{Congratulations}$ to Yat Sum from Cambourne Village College and Charlie Watkins from Bottisham Village College for successfully solving this problem!

Let $\angle BDC = \alpha$ . Since angles on a straight line add up to 180, $\angle ADE = 180^\circ - \angle BDC - \angle CBA = 180^\circ - 60^\circ - \alpha = 120^\circ - \alpha$ .

In $\triangle EAD$ and $\triangle DCB$ , the angles satisfy $\angle EAD = \angle DCB = 60^\circ$ and $\angle ADE = \angle CBD = 120^\circ - \alpha$ .

By the angle-angle similarity, it follows that $\triangle EAD \sim \triangle DCB$ .

From the similarity of the triangles, the corresponding sides are proportional:
$\frac{AE}{AD} = \frac{DC}{CB} = \frac{EB}{DB}$ .

Given that $DC = 12 \, \text{cm}$ and $CB = 48 \, \text{cm}$ , we have $\frac{DC}{CB} = \frac{12}{48} = \frac{1}{4}$ . Hence, $\frac{AE}{AD} = \frac{1}{4}$ .

$AD = AB - DC = 48 \, \text{cm} - 12 \, \text{cm} = 36 \, \text{cm}$ . Substituting this into the previous proportionality relationship yields,
$AE = \frac{AD}{4} = \frac{36 \, \text{cm}}{4} = 9 \, \text{cm}.$

Since $AB = AE + BE$ , $BE = AB - AE = 48 \, \text{cm} - 9 \, \text{cm} = 39 \, \text{cm}$ .

Thus, BE = 39cm.

Problem of the Week

Problem 14 Solution

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