Question

# D and E are points on the sides AB and AC respectively of a $∆ABC$. In each of the following cases, determine whether $DE\parallel BC$ or not. (i) $AD=5.7\mathrm{cm},DB=9.5\mathrm{cm},BD=4.8\mathrm{cm}\mathrm{and}EC=8\mathrm{cm}.$ (ii) $AB=11.7\mathrm{cm},AC=11.2\mathrm{cm},BD=6.5\mathrm{cm}\mathrm{and}AE=4.2\mathrm{cm}.$ (iii) $AB=10.8\mathrm{cm},AD=6.3\mathrm{cm},AC=9.6\mathrm{cm}\mathrm{and}EC=4\mathrm{cm}.$ (iv) $AD=7.2\mathrm{cm},AE=6.4\mathrm{cm},AB=12\mathrm{cm}\mathrm{and}AC=10\mathrm{cm}.$

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Solution

## (i) We have: $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{5.7}{9.5}=0.6\mathrm{cm}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{4.8}{8}=0.6\mathrm{cm}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\phantom{\rule{0ex}{0ex}}\mathrm{Applying}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{Thales}\text{'}\mathrm{theorem},\phantom{\rule{0ex}{0ex}}\mathrm{we}\mathrm{conclude}\mathrm{that}\mathrm{DE}\parallel \mathrm{BC}.$ (ii) We have: AB = 11.7 cm, DB = 6.5 cm Therefore, AD = 11.7 $-$ 6.5 = 5.2 cm Similarly, AC = 11.2 cm, AE = 4.2 cm Therefore, EC = 11.2 $-$ 4.2 = 7 cm $\mathrm{Now},\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{5.2}{6.5}=\frac{4}{5}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{4.2}{7}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\frac{\mathrm{AD}}{\mathrm{DB}}\ne \frac{\mathrm{AE}}{\mathrm{EC}}\phantom{\rule{0ex}{0ex}}$ Applying the converse of Thales' theorem, we conclude that DE is not parallel to BC. (iii) We have: AB = 10.8 cm, AD = 6.3 cm Therefore, DB = 10.8 $-$ 6.3 = 4.5 cm Similarly, AC = 9.6 cm, EC = 4 cm Therefore, AE = 9.6 $-$ 4 = 5.6 cm Now, $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{6.3}{4.5}=\frac{7}{5}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{5.6}{4}=\frac{7}{5}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\phantom{\rule{0ex}{0ex}}\mathrm{Applying}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{Thales}\text{'}\mathrm{theorem},\phantom{\rule{0ex}{0ex}}\mathrm{we}\mathrm{conclude}\mathrm{that}\mathrm{DE}\parallel \mathrm{BC}.$ (iv) We have: AD = 7.2 cm, AB = 12 cm Therefore, DB = 12 $-$ 7.2 = 4.8 cm Similarly, AE = 6.4 cm, AC = 10 cm Therefore, EC = 10 $-$ 6.4 = 3.6 cm Now, $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{7.2}{4.8}=\frac{3}{2}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{6.4}{3.6}=\frac{16}{9}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\frac{\mathrm{AD}}{\mathrm{DB}}\ne \frac{\mathrm{AE}}{\mathrm{EC}}\phantom{\rule{0ex}{0ex}}\mathrm{Applying}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{Thales}\text{'}\mathrm{theorem},\phantom{\rule{0ex}{0ex}}\mathrm{we}\mathrm{conclude}\mathrm{that}\mathrm{DE}\mathrm{is}\mathrm{not}\mathrm{parallel}\mathrm{to}\mathrm{BC}.$

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