To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides of $\frac{3}{7}$ the corresponding sides of $\bigtriangleup ABC$, first draw a ray BX such that CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B1, B2, B3, ... on BX at equal distances and next step is to join(A) B10 to C                (B) B3 to C                  (C) B7 to C                  (D) B4 to C

Solution Given: $\angle CBX$ is an acute angle.

Steps of construction
1.   Draw any ray BX making angle with BC on the side opposite to vertex A.
2.   Locate 7 points on BX in equidistant
3.   Now join Bto C

4.   Draw a line through B3${C}'$  parallel to B7C .

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