Answer:
Steps of construction (i) Draw a line segment \[BC=6\text{ }cm\]. (ii) With B as centre and radius equal to 5 cm, draw an arc. (iii) With C as centre and radius equal to 7 cm, draw an arc. (iv) Mark the point where the two arcs intersect as A. Join AB and AC Thus, \[\Delta \text{ }ABC\] is obtained. (v) Below BC, make an acute \[\angle CBX\]. (vi) Along BX, mark off five points \[{{B}_{1}},{{B}_{2}},{{B}_{3}},{{B}_{4}},{{B}_{5}}\] such that \[B\,{{B}_{1}}={{B}_{1}}{{B}_{2}}={{B}_{2}}{{B}_{3}}={{B}_{3}}{{B}_{4}}={{B}_{4}}{{B}_{5}}.\] (vii) Join \[{{B}_{5}}C\]. (viii) From \[{{B}_{4}}\] draw \[{{B}_{4}}D\parallel {{B}_{5}}C\], meeting BC at D. (ix) From D, draw \[DE\parallel CA\], meeting AB at E. Then, \[\Delta \,EBD\] is the required triangle each of whose sides is \[\frac{4}{5}\] of the corresponding side of \[\Delta \,ABC\].
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