Draw a triangle ABC with BC = 7 cm, . Then construct a triangle whose sides are times the corresponding sides of .

On the basis of the question, the figure can be drawn as follows:

We know that in a triangle sum of all its angles is 180^o

∴ In triangle ABC, we have

∠ A + ∠ B + ∠ C = 180^o

105^o + 45^o + ∠ C = 180^o

∠ C = 180^o – 150^o

∠ C = 30^o

Steps of construction:

(i) Firstly, draw a triangle ABC having side BC = 7 cm, ∠ B = 45^o and ∠ C = 30^o

(ii) Now draw a BX by making an acute angle with the side BC on the opposite side of the vertex A

(iii) After this locate 4 points on BX i.e. B₁, B₂, B₃ and B₄

(iv) Now, join B₃C and then draw a parallel line through B₄ to B₃C which intersect the extended BC at C’

(v) From C’ draw a line parallel to AC which intersects the extended line segment at C’

∴

Justification:

From the figure, we have:

As line A’C’ is parallel to Ac, therefore, it will make the same angle with the line BC

∴ ∠ A’C’B = ∠ ACB (Corresponding angles are equal) (i)

Now, in triangle A’B’C’ and triangle ABC we have:

∠ B = ∠ B (Common)

∠ A’C’B = ∠ ACB [From equation (i)]

∴ (By AA similarity rule)

As the corresponding sides of both the similar triangles are in the same ratio

∴

Hence, the construction is justified.