Question

to divide a line segment BC internally in the ratio 3:5 we draw a ray BX such that angle CBX is an acute angle .what will be the minimum number of point to be located at equal distance on ray BX​

Answer 1

8  points are required  To divide a line segment BC in the ratio 3:5

Step-by-step explanation:

To divide a line segment BC  in the ratio 3:5,

Step1 : Draw a line segment BC  of some length

Step 2 :  Draw a line segment BX such that ∠CBX is an acute angle

Step 3:  Take 8 point on BX of Equal length one by one ( consecutively)

Step 4 :  Join 8th Point with C as a straight line

Step 5 : Draw a line parallel to line drawn in step 4 such that it passes through 3rd point of step 3 and intersect BC at M

M divides BC in to  3 : 5 Ratio.

8 points are required

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Answer 2

Step-by-step explanation:

draw a line segment BC sum of length .

draw a line segment BX such that angle CBX is an acute angle .

take 8 points on BX of equal length one by one .

join 8th point C as a straight line .

draw a line parallel to line drawn in step 4 such that it passes through 3rd point of step 3 and intersect BC at M.

M divides BC in to 3:5 ratio