Step of Construction :

(i)    Draw a line segment BC = 8 cm.
(ii)   Draw a perpendicular bisector AD (4 cm) of BC.
(iii)  Joining AB and AC we get isosceles ΔABC.

(iv)    Construct an acute ∠CBX downwards.
(v)    Along BX mark off 3 equal points B₁, B₂, B₃ such that BB₁ = B₁B₂ = B₂B₃.
(vi)    Join C to B₂ and draw a line through B₃ parallel to B₂C intersecting the extended line segment BC at C’.
(vii)    Again draw a parallel line C’ A’ to AC cutting BP at A’.
(viii)    ΔA ‘BC’ is the required triangle.
Justification :
C’A’ || CA [By construction] ΔABC ~ ΔA ‘BC’
[Using AA similarity condition]

Steps of Construction :

(i) Construct a ΔABC in which AB = 6 cm, AC = 4 cm, BC = 5 cm.
(ii)    At A draw an acute ∠BAX below base AB.
(iii)    Along AX mark off points A₁, A₂, A₃ such that AA₁ = A₁ A₁ = A₂ A₃.
(iv)    Join A₃B.
(v)    From A₂ draw A₂B’ || A₃B meeting AB at B’.
(vi)    From B’ draw B‘C’ || CB meeting AC at C’. Thus, ΔAB‘C’ is the required triangle, each of whose sides is (2/3)rd of the corresponding sides of the ΔABC.
Justification :

[From (i)]
[By Basic proportionality theorem)

Steps of Construction :

(i)     Construct a ΔABC in which AB = 7cm, AC = 5 cm, BC = 6 cm.
(ii)    At A draw an acute ΔBAX below base AB.
(iii)   Along AX, mark off 7 points A₁, A₂, A₁,.,
A₇. Such that AA₁ = A₁ A₂ = A₂ A₃
= A₃ A₄ =......= A₆A₇.

(iv) Join A₅B.
(v) From A₇ draw A₇B’ || A₅B meeting produced part of AB at B’.
(vi) From B’, draw B‘C’ || BC intersecting the extended line segment AC at C’.
Thus, AB‘C’ is the required triangle each of whose sides is 7/5 of the corresponding sides of the triangle ΔABC.
Justification :

Steps of Construction :
(i)     Draw AB = 5.6 cm
(ii)    At a draw an acute ∠BAX below base AB.

(iii)    On AX make 5 + 8 i.e. 13 equal parts and mark them as A₁, A₂, A₃, A₄,... A₁₃

(iv)    Join B to A₁₃. From A₅ draw A₅C || A₁₃B. C is the required point of division and AC : CB = 5 : 8.

On measuring, we get
AC = 3.1 cm,
CB = 4.5 cm
Justification :

[Using basic proportionally theorem]



Therefore,    

This shows that C divides AB in the ratio 5 : 8.

Steps of Construction :

(i)    Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
(ii)    Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
(iii)    Along BX, mark off 4 points B₁, B₂, B₃ and B₄ on BX such that BB₁ = B₁B₂ = B₂ B₃ = B₃B₄.
(iv) Join B₄C and draw a line through B₃ parallel to B₄ C intersecting BC to C’.
(v) Draw a line through C’ parallel to the line CA to intersect BA at A’.



Then A ‘BC’ is the required triangle. Justification :


[By the Basic Proportionality Theorem]