Draw ABC in which AC = 6 cm, A = 45°, B = 105°. Then construct another triangle whose sides are 4/5 of the corresponding sides of ABC.

Given: ∠A = 45° and ∠B = 105°

We know that sum of all interior angles in a triangle is 180°.

∴ ∠A + ∠B + ∠C = 180°

⇒ 45° + 105° + ∠C = 180°

⇒ ∠C = 180° - 150°

⇒ ∠C = 30°

Construction Steps:

Step 1: Draw a ΔABC with side AC = 6cm, ∠A = 45° and ∠B = 30°.

Step 2: Draw any ray AX making an acute angle with AC on the side opposite to the vertex B.

Step 3: Locate 5 (the greater of 4 and 5 in 4/5) points A₁, A₂, A₃, A₄ and A₅ on AX so that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.

Step 4: Join A₅C and draw a line through A₃ (the 4th point, 4 being smaller of 4 and 5 in 4/5) parallel to A₅C to intersect AC at C′.

Step 5: Draw a line through C′ parallel to the line CB to intersect AB at B′.

Then, Δ AB’C′ is the required triangle.

Justification:

After dividing the line segment AC in the given ratio, AC’/C’C = 4/1

⇒ AC/AC’ = AC’ + C’C / AC’ = 1 + C’C/AC’ = 1 + 1/4 = 5/4

∴ BC/BC’ = 5/4

Also, C’B’ is parallel to CB.

∴ ΔAB’C’ ~ ΔABC

So, B’A/BA = B’C’/BC = AC’/AC = 4/5

This justifies the construction.