Given, a Δ ABC, we are required to construct a triangle whose side are of the corresponding sides of Δ ABC
1) Draw any ray BX making an acute angel with BC on the sides opposite to the vertex A.
2) Locate the points B1, B2, B3, B4, B5 on BX such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
3) Join B3 to C as 3 being smaller and through B5 draw a line parallel to B3C, intersecting the extended line segment BC at C’.
4) Draw a line through C parallel to CA intersecting the extended line segment BA at A’.
Then A’BC’ is the required triangle.
Note Δ ABC ∼ Δ A’BC’
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