, 2024

Solution 1:

Suppose that $BE = AC = x$ and $DE = y$. Extend $BC$ to point $F$ so that $BC = DE = y$.

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Since $BC + DE = 288$, then $BF = BC + CF = BC + DE = 288$. Also, $\triangle BED$ is congruent to $\triangle ACF$ by side-angle-side. Therefore, $$\angle BAF = \angle BAC + \angle ACF = \angle BDE + \angle DBE = 90\degree$$ since $DE$ and $AC$ are parallel. Next, $\triangle BED$ is similar to $\triangle BAF$ since both are right-angled and they share an angle at $B$. Therefore, $\$\dfrac{DE}{BD} = \dfrac{FA}{BF}$$and so$$\dfrac{DE}{120} = \dfrac{120}{288}$$, which gives$DE = $\text{120}$ 120}{288} =
50$, as required. Solution 2: Suppose that$BE = AC = x$and$DE = y$. [[IMAGE1]] Since$DE + BC = 288$, then$BC = 288 - y$. We note that$$\triangle$ BED$is similar to$$\triangle$ BCA because each is right-angled and their angles at

B$are common. Therefore,$$\dfrac{BE}{DE} = \dfrac{BC}{AC}$$and so$$\text{x y = 288}$ - y}{x}$.

Manipulating, we obtain $x^2 = y(288-y)$ and so $x^2 = 288y - y^2$ or $x^2 + y^2 = 288y$.

Also, using the Pythagorean Theorem in $\triangle BED$ gives $x^2 + y^2 = 120^2$. Since $x^2 + y^2 = 288y$ and $x^2 + y^2 = 120^2$, then $288y = 120^2$ which gives $2 \cdot 12 \cdot 12 \cdot y = 120 \cdot 120$ and so $2y = 10 \cdot 10$ or $y = 50$. Therefore, $DE = 50$.