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Basic Proportionality Theorem

In a Δ ABC, D...

Question

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.

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Solution

DE will be || to BC when $fraction numerator A D over denominator A B end fraction equals fraction numerator A E over denominator A C end fraction$ ( it shows that ABC and ADE are similar triangles with common angle A, Hence the other two corresponding angles will be equal too)

(i) $fraction numerator A D over denominator A B end fraction equals fraction numerator A E over denominator A C end fraction$

Hence DE|| BC

(ii) $fraction numerator A D over denominator A B space end fraction equals fraction numerator 1.4 over denominator 5.6 end fraction equals 1 fourth$
$fraction numerator A E over denominator A C space end fraction equals fraction numerator 1.8 over denominator 7.2 end fraction equals 1 fourth$
$fraction numerator A D over denominator A B end fraction equals fraction numerator A E over denominator A C end fraction$
Hence DE|| BC

(iii) $fraction numerator A D over denominator A B space end fraction equals fraction numerator 10.8 minus 4.5 over denominator 10.8 end fraction equals fraction numerator 6.3 over denominator 10.8 end fraction equals 7 over 12$
$fraction numerator A E over denominator A C space end fraction equals fraction numerator 2.8 over denominator 4.8 end fraction equals 7 over 12$
$fraction numerator A D over denominator A B end fraction equals fraction numerator A E over denominator A C end fraction$
Hence DE|| BC

(iv) $fraction numerator A D over denominator A B space end fraction equals fraction numerator 5.7 over denominator 5.7 plus 9.5 end fraction equals fraction numerator 5.7 over denominator 15.2 end fraction equals 3 over 8$
$fraction numerator A E over denominator A C space end fraction equals fraction numerator 3.3 over denominator 3.3 plus 5.5 end fraction equals 3 over 8$
$fraction numerator A D over denominator A B end fraction equals fraction numerator A E over denominator A C end fraction$
Hence DE|| BC

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Similar questions

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 .

Q. D and E are points on the sides AB and AC respectively of a

∆ A B C

. In each of the following cases, determine whether

D E ∥ B C

or not.

(i)

A D = 5.7 cm, D B = 9.5 cm, B D = 4.8 cm and E C = 8 cm .

(ii)

A B = 11.7 cm, A C = 11.2 cm, B D = 6.5 cm and A E = 4.2 cm .

(iii)

A B = 10.8 cm, A D = 6.3 cm, A C = 9.6 cm and E C = 4 cm .

(iv)

A D = 7.2 cm, A E = 6.4 cm, A B = 12 cm and A C = 10 cm .

D and E are points on the sides AB and AC respectively of a $Δ A B C$ . In each of the follwing cases, determine whether DE || BC or not.

(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm.

(ii) AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm and AE = 4.2 cm.

(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm and EC = 4 cm.

(iv) AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm and AC = 10 cm.

D

and

E

are respectively the points on the sides

A B

and

A C

of a

Δ A B C

such that

A B = 12 c m, A D = 8 c m, A E = 12 c m

and

A C = 18 c m

, show that

D E ∥ B C

D and E are respectively the points on the sides AB and AC of a $Δ$ ABC such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, prove that DE || BC.

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DE will be || to BC when ( it shows that ABC and ADE are similar triangles with common angle A, Hence the other two corresponding angles will be equal too) (i) Hence DE|| BC (ii) Hence DE|| BC (iii) Hence DE|| BC (iv) Hence DE|| BC