D and E are points on the sides AB and AC respectively of a $\triangle\text{ABC}.$ In the following cases, determine whether DE || BC or not. AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm.
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We have:
AD = 7.2cm, AB = 12cm
Therefore,
DB = 12 - 7.2 = 4.8cm
Similarly,
AE = 6.4cm, AC = 10cm
Therefore,
EC = 10 - 6.4 = 3.6cm
Now,
$\frac{\text{AD}}{\text{DB}}=\frac{7.2}{4.8}=\frac{3}{2}$
$\frac{\text{AE}}{\text{EC}}=\frac{6.4}{3.6}=\frac{16}{9}$
Thus, $\frac{\text{AD}}{\text{DB}}\not=\frac{\text{AE}}{\text{EC}}$
Applying the converse of Thalse' theorem, we conclude thet DE is not parallel to BC.
AD = 7.2cm, AB = 12cm
Therefore,
DB = 12 - 7.2 = 4.8cm
Similarly,
AE = 6.4cm, AC = 10cm
Therefore,
EC = 10 - 6.4 = 3.6cm
Now,
$\frac{\text{AD}}{\text{DB}}=\frac{7.2}{4.8}=\frac{3}{2}$
$\frac{\text{AE}}{\text{EC}}=\frac{6.4}{3.6}=\frac{16}{9}$
Thus, $\frac{\text{AD}}{\text{DB}}\not=\frac{\text{AE}}{\text{EC}}$
Applying the converse of Thalse' theorem, we conclude thet DE is not parallel to BC.
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