Answer
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Hint: Need to follow the construction steps in order.
Complete step-by-step answer:
Steps to construct the triangle:-
Draw a line segment BC = 6cm
At B, draw $\angle CBE = 60^\circ $
Draw $BF \bot BE$
Draw the perpendicular bisector of BC from point D, intersecting BF at O.
Draw a circle with O as centre and radius equal to OB or OC.
With D as centre and radius 8 cm, mark an arc which intersects BC produced in C’.
Draw a line parallel to AC at C’.
Extend the line segment BA, which intersect the line passing through C’ in A’.
Thus $\vartriangle A'BC'$ is the required triangle.
$\vartriangle A'BC' \sim \vartriangle ABC$ [AA similarity]
Note: We need to be particular while constructing the triangle. The measurements of line segments and angles should be perfect. AA similarity states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.
Complete step-by-step answer:
Steps to construct the triangle:-
Draw a line segment BC = 6cm
At B, draw $\angle CBE = 60^\circ $
Draw $BF \bot BE$
Draw the perpendicular bisector of BC from point D, intersecting BF at O.
Draw a circle with O as centre and radius equal to OB or OC.
With D as centre and radius 8 cm, mark an arc which intersects BC produced in C’.
Draw a line parallel to AC at C’.
Extend the line segment BA, which intersect the line passing through C’ in A’.
Thus $\vartriangle A'BC'$ is the required triangle.
$\vartriangle A'BC' \sim \vartriangle ABC$ [AA similarity]
Note: We need to be particular while constructing the triangle. The measurements of line segments and angles should be perfect. AA similarity states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.
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