Can somebody plz give me the steps of construction for the following sum:- ?
Construct ΔABC in which BC = 4.2 cm, ∠ABC = 60° and AB = 5 cm. Construct a circle of radius 2 cm to touch both the arms of ∠ABC of ΔABC.
Draw AX of sufficient length. Taking 5cm radius in compass, cut off AX at B . Now AB =5cm.
Take the compass with 5cm and at A as centre, make an arc and similarly with the same radius in compass and B as centre and make arc. The two arc cut at a point P. Draw BP and mark off C on BP such that BC = 4.2 cm. Now Angle ABC = 60 degree.
Bisect the angle ABC : With B as centre and with suitable radius,draw an arc to cut BA and BC at X and Y respectively.
With X as centre and then Y as centre , draw arcs of equal radius to cut at Z. Draw BZ , the angle bisector of angle ABC. Extend BZ to a sufficient length.
Mark off BO along BZ, such that BO = 4cm.
Draw a circle with on BO as diameter with the centre L , the mid point of BO with radius LO= LB =2cm.
The circle cuts the BA and BC at Points M and N.
Then OM = ON should be equal to 2cm. And with O as centre you can draw a circle of radius OM =ON = 2cm , which touch both AB and BC at M and N.
Proof: Since BZ is bisector of angle ABC, Angle ABZ = Angle ZBC = 30 degree. Triangle OMB = is a right angled triangle as it is in a semimicirlce. Therefore, angle OMB = 90 degree. Angle OBM= angle OBA = 30 degree. Therefore, in a right angled triangle , where one angle is 30 degree, the opposite side OM must be half of hypotenuse BO. So OM =2cm. Since angle BMO =90 degree, BA is the tangent to the circle of radius 2cm with centre O and BA touches the circle at M. Similar explanation could be given to say that BC is the tangent to the circle touching at N.