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# Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $$\frac{2}{3}$$ of the corresponding sides of the first trianngle.

Steps of Construction from the given data is done in the following steps :
Step 1. Draw a line segment BC=6 cm.

Step 2. With B as a centre draw an arc of radius equal to 5 cm.

Step 3. With C as centre draw an arc of radius equal to 4 cm, such that this arc intersects the previously drawn arc at A.

Step 4. Join both AB and AC,such that we get ΔABC, which is our required triangle.

Step 5. Below the line segment BC, make an acute angle CBX.

Step 6. Along BX, locate three points: $$B_1,B_2 and B_3$$such that $$BB_1=B_1B_2=B_2B_3$$

Step 7. Join $$B_3C$$ Step 8. From $$B_2$$, draw a line $$B_2D$$ ∥ $$B_3C$$such that it meets BC at D.

Step 9. From D, draw ED ∥AC such that it meets BA at E. Then, EBD is the required triangle whose sides are $$\frac{2}3$$ rd of the corresponding sides of ΔABC.

JUSTIFICATION:

By construction we have

$$\frac{BD}{DC}=\frac{2}1$$

Therefore, $$\frac{BC}{BD}=\frac{BD+DC}{BD}=1+\frac{DC}{BD}$$

$$\frac{BC}{BD}=\frac{3}2 so \frac{BD}{BC}=\frac{2}3$$

Also we have DE and AC as parallel

Therefore, ΔABC~ΔEBD(By AA property)

{$$\angle$$ D=$$\angle$$ C (by construction) and $$\angle$$ A is common}

So $$\frac{EB}{AN}=\frac{BD}{BC}=\frac{DE}{CA}=\frac{2}3$$

Hence, we get the new triangle whose sides are equal to $$\frac{2}{3}$$ rd of the corresponding sides of $$\triangle$$ ABC