Question

# Construct a $$\triangle ABC$$ in which $$AB = 4\ cm,\ BC =5\ cm$$ and $$AC = 6\ cm.$$ Now, construct a triangle similar to $$\triangle ABC$$ such that each of its sides is two-third of the corresponding sides of $$\triangle ABC.$$ Also, prove your assertion.

Solution

## Steps:1. draw base $$AB=4cm$$2. with A as center, with a scale of 5cm as radius draw an arc and with B as center, with a scale of 6cm as radius draw an arc.3. let C be the point of intersection of the above 2 lines.4. draw a ray AX making an acute angle with line AB on the opposite side if the vertex C. 5. mark 3 points $$A1,A2,A3$$ such tha $$A1A2=A2A3=A3A1$$6. join $$A3B$$ and draw a parallel line $$B'A2$$7. draw a line parallel to $$BC$$  to intersect $$AC$$ at $$C'$$Thus $$AB'C'$$ is a required triangle.Justification;By construction,$$\dfrac{AB'}{AB}=\dfrac{AA2}{AA3}=\dfrac{2}{3}$$also $$B'C'\parallel BC$$$$\therefore \angle AB'C'=\angle ABC$$In $$\triangle AB'C',\triangle ABC$$$$\angle A=\angle A$$$$\angle AB'C'=\angle ABC$$$$\triangle AB'C'\sim \triangle ABC$$$$\Rightarrow \dfrac{AB'}{AB}=\dfrac{AC'}{AC}=\dfrac{B'C'}{BC}$$$$\therefore \dfrac{AB'}{AB}=\dfrac{AC'}{AC}=\dfrac{B'C'}{BC}=\dfrac{2}{3}$$Maths

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