Steps of construction:
1. Draw a line segment AB=5 cm.
2. From point B, draw ∠ABY=60°.
To measure angle at B:
a. With B as centre and with any radius, draw another arc cutting the line AB at D.
b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point E.
c. Draw a ray BY passing through E which forms an angle of 60° with the line AB.
3. With B as centre and radius of 6 cm draw an arc intersecting the line BY at C.
4. Join AC, ∆ABC is the required triangle.
5. From A, draw any ray AX downwards making an acute angle.
6. Mark 7 points A1,A2,A3,A4,A5,A6 and A7 on AX such that AA1=A1A2=A2A3=A3A4=A4A5=A5A6=A6A7
7. Join A7B and from A5 draw A5M||A7B intersecting AB at M.
8. From point M draw MN││BC intersecting AC at N. Then, ∆AMN is the required triangle whose sides are equal to of the corresponding sides of ∆ABC.
Let AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = x
As per the construction A5M││A7B
Also MN || BC,
Therefore rAMC is congruent to rAMN
Hence, the new triangle rAMN is similar to the given triangle rABC and its sides are times of the corresponding sides of rABC.
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