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) $AD=7.2\mathrm{cm},AE=6.4\mathrm{cm},AB=12\mathrm{cm}\mathrm{and}AC=10\mathrm{cm}.$

Answer 1

(i) We have:

$$\begin{gathered} \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{5.7}{9.5}=0.6\mathrm{cm} \\ \frac{\mathrm{AE}}{\mathrm{EC}}=\frac{4.8}{8}=0.6\mathrm{cm} \\ \mathrm{Hence},\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}} \\ \mathrm{Applying}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{Thales}\text{'}\mathrm{theorem}, \\ \mathrm{we}\mathrm{conclude}\mathrm{that}\mathrm{DE}\parallel \mathrm{BC}. \end{gathered}$$

(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

$$\begin{gathered} \mathrm{Now}, \\ \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{5.2}{6.5}=\frac{4}{5} \\ \frac{\mathrm{AE}}{\mathrm{EC}}=\frac{4.2}{7} \\ \mathrm{Thus},\frac{\mathrm{AD}}{\mathrm{DB}}\ne \frac{\mathrm{AE}}{\mathrm{EC}} \end{gathered}$$

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,
$$\begin{gathered} \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{6.3}{4.5}=\frac{7}{5} \\ \frac{\mathrm{AE}}{\mathrm{EC}}=\frac{5.6}{4}=\frac{7}{5} \\ \Rightarrow \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}} \\ \mathrm{Applying}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{Thales}\text{'}\mathrm{theorem}, \\ \mathrm{we}\mathrm{conclude}\mathrm{that}\mathrm{DE}\parallel \mathrm{BC}. \end{gathered}$$

(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,
$$\begin{gathered} \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{7.2}{4.8}=\frac{3}{2} \\ \frac{\mathrm{AE}}{\mathrm{EC}}=\frac{6.4}{3.6}=\frac{16}{9} \\ \mathrm{Thus},\frac{\mathrm{AD}}{\mathrm{DB}}\ne \frac{\mathrm{AE}}{\mathrm{EC}} \\ \mathrm{Applying}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{Thales}\text{'}\mathrm{theorem}, \\ \mathrm{we}\mathrm{conclude}\mathrm{that}\mathrm{DE}\mathrm{is}\mathrm{not}\mathrm{parallel}\mathrm{to}\mathrm{BC}. \end{gathered}$$