In the adjoining figure, AD=5.6 cm, AB=8.4cm, AE=3.8cm and AC=5.7cm. Show that DE||BC.

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Solution

We have,
AD=5.6cm, DB=(AB-AD)=(8.4-5.6)CM=2.8CM
AE=3.8cm, EC=(AC-AE)=(5.7-3.8)CM=1.9CM
$∴ \frac{A E}{E C} - \frac{3.8}{1.9} - \frac{2}{1} a n d \frac{A D}{D B} - \frac{5.6}{2.8} - \frac{2}{1}$
Thus, $\frac{A D}{D B} - \frac{A E}{E C}$
$∴$ DE divides AB and AC proportionally.
Hence, DE||BC

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Similar questions

In the adjoining figure, AD=5.6 cm, AB=8.4cm, AE=3.8cm and AC=5.7cm. Show that DE||BC.

In the adjoining figure, $∠ A B C = 90^{\circ}$ and $B D ⊥ A C$ . If AB = 5.7 cm, BD = 3.8 cm, CD = 5.4 cm, then find the length of BC.

△

A B C, D

E

D E

∥

B C

A B = 12 c m, A D = 8 c m, A E = 12 c m a n d A C = 18 c m

A B = 5.6 c m, A D = 1.4 c m, A C = 7.2 c m a n d A E = 1.8 c m

A B = 10.8 c m, B D = 4.5 c m, A C = 4.8 c m a n d A E = 2.8 c m

A D = 5.7 c m, B D = 9.5 c m, A E = 3.3 c m a n d E C = 5.5 c m .

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