Discussion
ABC is a right triangle, right-angled at B. D and E are points on side AB such that AD = DE = EB. Then, \(\dfrac{DC^{2} - BC^{2}}{AC^{2} - EC^{2}}\) is equal to :
Answer: D
The ratio is \(\dfrac{1}{2}\).
Let \(AD = DE = EB = a\), so \(AB = 3a\), \(DB = 2a\) and \(EB = a\). Because the angle at B is a right angle, Pythagoras' theorem applies to triangles DBC, ABC and EBC:
\[ \begin{aligned} DC^{2} - BC^{2} &= DB^{2} = 4a^{2} \\ AC^{2} - EC^{2} &= AB^{2} - EB^{2} \\ &= 9a^{2} - a^{2} = 8a^{2} \end{aligned} \]
\[ \frac{4a^{2}}{8a^{2}} = \frac{1}{2} \]
BC cancels in both differences, so the answer does not depend on the length of BC. \(\frac{4}{9}\) is \(DB^{2} : AB^{2}\), which forgets to subtract \(EB^{2}\).
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