In the following question how can we draw the altitude?

Construct a triangle ABC with BC=4.2cm, angle B=45 deg. and altitude through A is 2.5 cm. Draw another triangle similar to this triangle with scale factor ½.

Consider the rough figure of required triangle ABC.

Given : ∠B = 45°

AD is the altitude through A.

∴ AD = 2.5 cm and ∠ADB = 90°

In ΔADB,

∠B + ∠ADB + ∠BAD = 180°

⇒ 45° + 90° + ∠BAD = 180°

 ∠BAD = 45°

This implies, ΔADB is an isosceles triangle.

∴ AD = BD = 2.5 cm

Step of construction:

1. Draw a line segment BC = 4.2 cm.

2. At B construct an angle of 45° i.e. ∠XBC = 45°.

3. From point B, and a radius of 2.5 cm, draw an arc on BC, intersecting at D on BD.

4. Construct perpendicular at D, such that YD intersecting BX at A.

5. Join A to C. ΔABC in the required triangle.

Steps to draw similar triangle sealing factor .

1. Draw and ray BZ, making and acute angle with BC.

2. Locate 3 (= 1 + 2) points B₁ , B₂ and B₃ on BZ so that BB₁ = B₁B₂ = B₂B₃.

3. Join CB₃

4. Through the point B₁, draw a line parallel to CB₃ to intersect BC at C'.

5. Draw a line through C', parallel to the AC to intersect AB at A'. ΔA'BC' in the required triangle.