# Construct a trian

Steps of constructions:

1. Draw a line segment BC = 7 cm, and draw YBC = 45°.

2. Draw ZCB = 30° ; and let CZ and BY intersect at A. Therefore, ΔABC is ready.

[ C = 30°, as A = 105° ; B = 45° and we know, By angle sum property of triangle,

A + B + C = 180°

45° + 105° + C = 180°

C = 30° ]

3. Below BC, make an acute angle CBX.

4. Along BX mark 4 (greater of 3 and 4 in 3/4) points B1, B2, B3,B4 at equal distances.

5. Join B4C

6. From point B3, make a line B3C' parallel to B4C , meeting BC at C' [Choosing B3 because we need a 3:4 ratio]

7. From point B draw a line C'A' parallel to CA.

8.A'BC' is the triangle, whose sides are three - fourth of ∆ABC

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