# Draw a triangle ABC with side BC=6cm,AB=5cm & angle ABC=60' . Then construct a triangle with sides 3/4 of the corresponding sides of the triangle ABC.

A ΔA'BC' whose sides are of the corresponding sides of ΔABC can be drawn as follows.

Step 1

Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 3 and 4), B1, B2, B3, B4, on line segment BX.

Step 4

Join B4C and draw a line through B3, parallel to B4C intersecting BC at C'.

Step 5

Draw a line through C' parallel to AC intersecting AB at A'. ΔA'BC' is the required triangle.

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Justification

The construction can be justified by proving

In ΔA'BC' and ΔABC,

∠A'C'B = ∠ACB (Corresponding angles)

∠A'BC' = ∠ABC (Common)

∴ ΔA'BC' ∼ ΔABC (AA similarity criterion)

… (1)

In ΔBB3C' and ΔBB4C,

∠B3BC' = ∠B4BC (Common)

∠BB3C' = ∠BB4C (Corresponding angles)

∴ ΔBB3C' ∼ ΔBB4C (AA similarity criterion)

From equations (1) and (2), we obtain

⇒

This justifies the construction.

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do we need to give justification also

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ssss
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Nufaila is write
I am satisfy this answer
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This is real Answer

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Pretty good question,Farina
pretty good, answer Nufaila

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