Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle
Solution:

Draw the line segment of the largest length 7 cm. Measure 5 cm and 6 cm separately and cut arcs from 2 ends of the line segment such that they cross each other at one point. Connect this point from both the ends

Then draw another line that makes an acute angle with the given line (7 cm). Divide the line into m + n parts where m and n are the ratios given.

Two triangles are said to be similar if their corresponding angles are equal are said to satisfy AngleAngleAngle (AAA) Axiom.

The basic proportionality theorem states that “If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally".
Steps of construction:
 Draw BC = 7cm with B and C as centres and radii 5 cm and 6 cm respectively. Draw arcs to intersect at A. ΔABC is obtained.
 Draw ray BX making ∠CBX acute.
 Mark 7 points (greater of 7 and 5 in 7/5 ) B_{1}, B_{2},………B_{7} on BX such that BB_{1} = B_{1}B_{2} =............... = B_{6}B_{7}
 Join B_{5} (smaller of 7 and 5 in 7/5 and so the 5th point) to C and draw B_{7}C' parallel to B5C intersecting the extension of BC at C'.
 Through C' draw C'A' parallel to CA to meet the extension of BA at A’. Now, ΔA' B'C' is the required triangle similar to ΔABC where BA'/BA = C'A'/CA = BC'/BC = 7
Proof:
In ΔBB_{7}C', B_{3}C is parallel to B_{7}C'
Hence by Basic proportionality theorem,
B_{6}B_{7}/BB_{5} = CC'/BC = 2/5
Adding 1,
CC'/BC + 1 = 2/5 + 1
(BC + CC')/BC = 7/5
BC'/BC = 7/5
Consider ΔBAC and ΔBA'C'
∠ABC = ∠A'BC' (Common)
∠BCA = ∠BC'A' (Corresponding angles ∵ CA  C'A')
∠BAC = ∠BA'C' (Corresponding angles)
By AAA axiom, ΔBAC ∼ ΔBA'C'
∴ Corresponding sides are proportional
Hence,
BA'/BA = C'A'/A = BC'/BC = 7/5
Video Solution:
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle
NCERT Solutions Class 10 Maths  Chapter 11 Exercise 11.1 Question 3:
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle
A triangle BAC of sides 5 cm, 6 cm, and 7 cm and another triangle B'A'C of sides 7/5 of the corresponding ones of triangle BAC have been constructed