To construct a triangle similar to a given ΔABC with its sides 3⁄7 of the corresponding sides of ΔABC, 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
B10 to C
B3 to C
B7 to C
B4 to C
We want a triangle whose sides are 3/7 of ΔABC.
Step 1: Draw a ray BX making an acute angle at B on the side opposite to vertex A.
Step 2: Mark 7 equal points on this ray (because the denominator is 7). Name them B₁, B₂, B₃, …, B₇.
Step 3: Since we want 3/7 of the original, the numerator is 3. So we need B₃.
Step 4: Join B₇ to C (the last division point to vertex C).
Step 5: Draw a line through B₃ parallel to B₇C. This will meet BC at a new point C′.
Step 6: Similarly, draw a line through C′ parallel to AC. This meets AB at A′.
Thus, ΔA′BC′ is the required triangle similar to ΔABC with sides 3/7 of it.
Correct join in the step asked: B3 to C.