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, DE || BC.
(i) If AD=3.4 cm, AB=8.5 cm and AC=13.5 cm, find AE.
(ii) If $\frac{A D}{D B} = \frac{3}{5}$ and AC=9.6 cm, find AE.

Q. In a

△

A B C, D

and

E

are points on the sides AB and AC respectively. For each of the following cases show that

D E

∥

B C

(i)

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

(ii)

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

(iii)

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

(iv)

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 .

In a $Δ A B C$ , D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC:
(i) AB = 12cm, AD = 8 cm, AE = 12 cm and AC= 18cm.
(ii)AB = 5.6cm, AD= 1.4cm, AC = 7.2 cm and AE= 1.8 cm.
(iii) AB= 10.8 cm, BD= 4.5cm, AC = 4.8cm and AE =2.8 cm.
(iv)AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and EC = 5.5 cm.

Q. In given figure