# Draw a triangle ABC with side BC = 6 cm, AB = 5cm, and ∠ABC = 60°, then construct a triangle whose sides are of the corresponding sides of ABC.

Step1: Construct segment BC of 6 cm

Step2: construct a ray at angle 60° above BC from point B

Step3: take 5 cm in compass because of AB = 5 cm, keep the needle of the compass on point B and mark an arc intersecting the ray drawn in step 2. Mark the intersection point as A and join A and C hence ΔABC is ready

Step4: draw a ray at any angle from point B below BC

Step5: take any distance in compass and keeping the needle of the compass on point A cut an arc on ray constructed in step4 and name that point X1. Keeping the distance in compass same keep the needle of the compass on point X1 and cut an arc on the same ray and mark that point as X2. Draw 4 such parts (greater of 3 and 4 in 3/4), i.e. by repeating this process mark points upto X4

Step6: join X4 and C and from X3 (because X3 is the third point 3 being smaller in 3/4) construct line parallel to X3C and mark the intersection point with BC as D

BD is three forth of BC

Step7: construct line parallel to AC from point D and mark the intersection point with AB as E thus ΔBDE which have sides three forth of ΔABC is ready.

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