Megha draws and angle ∠ABC = 60° by a scale and a compass. Let’s take two pints P and Q on the rays BA and BC respectively. Let’s draw a straight line through the point P parallel to the ray BC and also draw a straight line through the point Q parallel to the ray BA.

Let D be the intersecting Point. Let’s write the type of quadrilateral that PBQD is.

Let us first construct what Megha had drawn

∠ABC = 60° only using scale and compass

Step1: Draw a ray BC. Take any distance in compass keep the needle of compass on point B mark an arc intersecting BC at X

Step2: Keeping the same distance in compass as that in step1 keep the needle on point X and mark an arc intersecting the arc drawn in step1 at Y. Draw ray AB passing though Y ∠ABC = 60° is ready

Now the quadrilateral part

Step3: Take any points P and Q on BA and BC respectively

Constructing a line parallel to BC passing through P

Step4: Take the same distance in compass as that of in step1. Keep the needle on point P and mark an arc intersecting PA at M

Step5: Take distance XY in compass keep the needle of compass on point M and draw an arc intersecting arc drawn in step4 at N. Draw line passing through PN

Constructing a line parallel to BA passing through Q

Step6: Take the same distance in compass as that of in step1. Keep the needle on point Q and mark an arc intersecting QC at R

Step7: Take distance XY in compass keep the needle of compass on point R and draw an arc intersecting arc drawn in step6 at S. Draw line passing through QS which intersects line PN at D

From figure we have constructed

PD || BQ and BP || QD opposite sides of quadrilateral PBQD are parallel hence PBQD is a parallelogram